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  4. Direct-current circuits
  5. Kirchhoff's rules

10.3 Kirchhoff's rules  (Page 7/10)

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Solar cell arrays

Another example dealing with multiple voltage sources is that of combinations of solar cells —wired in both series and parallel combinations to yield a desired voltage and current. Photovoltaic generation, which is the conversion of sunlight directly into electricity, is based upon the photoelectric effect. The photoelectric effect is beyond the scope of this chapter and is covered in Photons and Matter Waves , but in general, photons hitting the surface of a solar cell create an electric current in the cell.

Most solar cells are made from pure silicon. Most single cells have a voltage output of about 0.5 V, while the current output is a function of the amount of sunlight falling on the cell (the incident solar radiation known as the insolation). Under bright noon sunlight, a current per unit area of about 100 mA/cm 2 of cell surface area is produced by typical single-crystal cells.

Individual solar cells are connected electrically in modules to meet electrical energy needs. They can be wired together in series or in parallel—connected like the batteries discussed earlier. A solar-cell array or module usually consists of between 36 and 72 cells, with a power output of 50 W to 140 W.

Solar cells, like batteries, provide a direct current (dc) voltage. Current from a dc voltage source is unidirectional. Most household appliances need an alternating current (ac) voltage.

Summary

  • Kirchhoff’s rules can be used to analyze any circuit, simple or complex. The simpler series and parallel connection rules are special cases of Kirchhoff’s rules.
  • Kirchhoff’s first rule, also known as the junction rule, applies to the charge to a junction. Current is the flow of charge; thus, whatever charge flows into the junction must flow out.
  • Kirchhoff’s second rule, also known as the loop rule, states that the voltage drop around a loop is zero.
  • When calculating potential and current using Kirchhoff’s rules, a set of conventions must be followed for determining the correct signs of various terms.
  • When multiple voltage sources are in series, their internal resistances add together and their emfs add together to get the total values.
  • When multiple voltage sources are in parallel, their internal resistances combine to an equivalent resistance that is less than the individual resistance and provides a higher current than a single cell.
  • Solar cells can be wired in series or parallel to provide increased voltage or current, respectively.

Conceptual questions

Can all of the currents going into the junction shown below be positive? Explain.

The figure shows a junction with three incoming current branches.
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Consider the circuit shown below. Does the analysis of the circuit require Kirchhoff’s method, or can it be redrawn to simplify the circuit? If it is a circuit of series and parallel connections, what is the equivalent resistance?

The figure shows a circuit with positive terminal of voltage source V connected to three parallel branches. The first branch has resistor R subscript 2 connected to parallel branches with R subscript 4 and R subscript 3 series with R subscript 5. The second branch has resistor R subscript 1 and third branch has resistor R subscript 6.

It can be redrawn.
R eq = [ 1 R 6 + 1 R 1 + 1 R 2 + ( 1 R 4 + 1 R 3 + R 5 ) −1 ] −1

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Do batteries in a circuit always supply power to a circuit, or can they absorb power in a circuit? Give an example.

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What are the advantages and disadvantages of connecting batteries in series? In parallel?

In series the voltages add, but so do the internal resistances, because the internal resistances are in series. In parallel, the terminal voltage is the same, but the equivalent internal resistance is smaller than the smallest individual internal resistance and a higher current can be provided.

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Semi-tractor trucks use four large 12-V batteries. The starter system requires 24 V, while normal operation of the truck’s other electrical components utilizes 12 V. How could the four batteries be connected to produce 24 V? To produce 12 V? Why is 24 V better than 12 V for starting the truck’s engine (a very heavy load)?

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Problems

Consider the circuit shown below. (a) Find the voltage across each resistor. (b)What is the power supplied to the circuit and the power dissipated or consumed by the circuit?

The figure shows positive terminal of voltage source V subscript 1 of 12 V connected in series to resistor R subscript 1 of 10 kΩ connected in series to resistor R subscript 2 of 20 kΩ connected in series to resistor R subscript 3 of 10 kΩ connected in series to positive terminal of voltage source V subscript 2 of 24 V connected in series to resistor R subscript 4 of 10 kΩ connected in series to resistor R subscript 5 of 10 kΩ.
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Consider the circuits shown below. (a) What is the current through each resistor in part (a)? (b) What is the current through each resistor in part (b)? (c) What is the power dissipated or consumed by each circuit? (d) What is the power supplied to each circuit?

Part a shows positive terminal of voltage source V subscript 1 of 1.6 V connected to parallel branches, one with resistor R subscript 1 of 2 kΩ and second with positive terminal of voltage source V subscript 2 of 1.4 V and resistor R subscript 3 of 1 kΩ. The two branches are connected back to V subscript 1 through resistor R subscript 2 of 1 kΩ. Part b shows the same circuit as part a but the terminals of V subscript 2 are reversed.

a. I 1 = 0.6 mA , I 2 = 0.4 mA , I 3 = 0.2 mA ;
b. I 1 = 0.04 mA , I 2 = 1.52 mA , I 3 = −1.48 mA ; c. P out = 0.92 mW , P out = 4.50 mW ;
d. P in = 0.92 mW , P in = 4.50 mW

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Consider the circuit shown below. Find V 1 , I 2 , and I 3 .

The positive terminal of voltage source V subscript 1 is connected to resistance R subscript 1 of 12 Ω with right current I subscript 1 of 2 A connected to two parallel branches, first with resistor R subscript 2 of 6 Ω with upward current I subscript 2 and second with right current I subscript 3, negative terminal of voltage source V subscript 2 of 21 V and resistor R subscript 3 of 5 Ω.
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Consider the circuit shown below. Find V 1 , V 2 , and R 4 .

The figure shows a circuit with three horizontal branches. The first branch has resistor R subscript 1 of 6 Ω with right current I subscript 1 of 4 A. The second branch has resistor R subscript 2 of 4 Ω with left current I subscript 2 of 3 A and resistor R subscript 3 of 6 Ω with left current I subscript 3 of 1 A. The third branch has resistor R subscript 5 of 4 Ω with left current I subscript 5 of 3 A. The first and second horizontal branches are connected on the right directly and on the left with voltage source V subscript 1 with positive terminal connected to first branch. The second and third horizontal branches are connected on the right directly and on the left with resistor R subscript 4 with upward current I subscript 4 of 1 A. The second and third branches are also connected in the middle with a voltage source V subscript 2 with positive terminal connected to second branch.

V 1 = 42 V , V 2 = 6 V , R 4 = 6 Ω

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Consider the circuit shown below. Find I 1 , I 2 , and I 3 .

The positive terminal of voltage source V subscript 1 of 24 V is connected to two parallel branches. The first branch has resistor R subscript 1 of 8 Ω with downward current I subscript 1 and second branch connects to positive terminal of voltage source V subscript 2 of 10 V and resistor R subscript 3 of 4 Ω with left current I subscript 3. The two branches are connected to V subscript 1 through resistor R subscript 2 of 6 Ω with left current of I subscript 2.
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Consider the circuit shown below. (a) Find I 1 , I 2 , I 3 , I 4 , and I 5 . (b) Find the power supplied by the voltage sources. (c) Find the power dissipated by the resistors.

The circuit has four vertical branches. From left to right, first branch has voltage source V subscript 1 of 12 V with positive terminal upward. The second branch has resistor R subscript 1 of 4 Ω with downward current I subscript 1. The third branch has voltage source V subscript 2 of 5 V with positive terminal upward and upward current I subscript 5. The fourth branch has resistor R subscript 4 of 2 Ω with downward current I subscript 4. The first and second branch are connected at the bottom through resistor R subscript 2 of 3 Ω with left current I subscript 2 and second and third branch are connected at the bottom through resistor R subscript 3 of 2 Ω with left current I subscript 3.

a. I 1 = 1.5 A , I 2 = 2 A , I 3 = 0.5 A , I 4 = 2.5 A , I 5 = 2 A ; b. P in = I 2 V 1 + I 5 V 5 = 34 W ;
c. P out = I 1 2 R 1 + I 2 2 R 2 + I 3 2 R 3 + I 4 2 R 4 = 34 W

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Consider the circuit shown below. Write the three loop equations for the loops shown.

The circuit has four vertical branches. From left to right, first branch has voltage source V subscript 1 with positive terminal upward. The second branch has resistor R subscript 2 with downward current I subscript 2. The third branch has voltage source V subscript 2 with positive terminal upward and downward current I subscript 2. The fourth branch has resistor R subscript 5 with downward current I subscript 5. The first and second branch are connected at the bottom through resistor R subscript 1 and second and third branch are connected at the bottom through resistor R subscript 4 with left current I subscript 4. The second and third branch are connected at the top through resistor R subscript 3 with left current I subscript 3. The current at the top between first and second branch is right I subscript 1.
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Consider the circuit shown below. Write equations for the three currents in terms of R and V .

The circuit has four vertical branches. From left to right, first branch has voltage source V subscript 1 with positive terminal upward and resistor R. The second branch has resistor R with downward current I subscript 1. The third and fourth branches both have resistor 2 R and are connected to positive terminal of another voltage source V. The current between first and second branch is right I subscript 2 and between second and third branch is left I subscript 3.

I 1 = 3 5 V R , I 2 = 2 5 V R , I 3 = 1 5 V R

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Consider the circuit shown in the preceding problem. Write equations for the power supplied by the voltage sources and the power dissipated by the resistors in terms of R and V .

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A child’s electronic toy is supplied by three 1.58-V alkaline cells having internal resistances of 0.0200 Ω in series with a 1.53-V carbon-zinc dry cell having a 0.100 - Ω internal resistance. The load resistance is 10.0 Ω . (a) Draw a circuit diagram of the toy and its batteries. (b) What current flows? (c) How much power is supplied to the load? (d) What is the internal resistance of the dry cell if it goes bad, resulting in only 0.500 W being supplied to the load?

a.
The resistor R subscript L is connected in series with resistor r subscript 2, voltage source ε subscript 2, resistor r subscript 1, voltage source ε subscript 1, resistor r subscript 1, voltage source ε subscript 1, resistor r subscript 1and voltage source ε subscript 1. All voltage sources have upward negative terminals. ;
b. 0.617 A; c. 3.81 W; d. 18.0 Ω

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Apply the junction rule to Junction b shown below. Is any new information gained by applying the junction rule at e ?

The circuit has three vertical branches. From left to right, first branch has voltage source ε subscript 1 of 18 V and internal resistance 0.5 Ω with positive terminal upward. The second branch has resistor R subscript 2 of 6 Ω with downward current I subscript 3 and voltage source ε subscript 2 of 3 V and internal resistance 0.25 Ω with positive terminal downward. The third branch has voltage source ε subscript 3 of 12 V and internal resistance 0.5 Ω with positive terminal downward. The first and second branch are connected at the top through resistor R subscript 1 of 20 Ω with right current I subscript 1 and bottom through resistor R subscript 4 of 15 Ω. The second and third branch are connected at the top through resistor R subscript 3 of 8 Ω with right current I subscript 2 and bottom through voltage source ε subscript 4 of 18 V with right positive terminal and internal resistance 0.75 Ω.
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Apply the loop rule to Loop afedcba in the preceding problem.

I 1 r 1 − ε 1 + I 1 R 4 + ε 4 + I 2 r 4 + I 4 r 3 − ε 3 + I 2 R 3 + I 1 R 1 = 0

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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