## Mastering Physics Solutions: Batteries in Series or Parallel

Batteries in Series or Parallel

Part A = counterclockwise
Part B = I = 2EMF / ((2 * R1) + R2)
Part C = IB = 2*EMF / (R1 + 2*R2)
Part D = 0.064 W
Part E = R1 / R2 = 1
Part F = R2 > R1

Solution Below:

You are given two circuits with two batteries of emf Ε and internal resistance R1 each. Circuit A has the batteries connected in series with a resistor of resistance R2, and circuit B has the batteries connected in parallel to an equivalent resistor.

Note that the symbol Ε should be entered in your answers as EMF.

Part A

In which direction does the current in circuit A flow?

Remember that current flows in the direction of the battery (from the short line to the long line, on the diagram). So the current flows counterclockwise.

counterclockwise

Part B

What is the current through the resistor of resistance R2 in circuit A?
Express the current in terms of EMF, R1, and R2.

Remember that V = IR. Also remember that the current everywhere in a circuit is the same. So all you need to do is find the current according to the formula V = IR.

V = 2 * EMF (since there are 2 batteries). So:

2 * EMF = I * R

Since the resistors are in series, add them together:

2 * EMF = I * ((2 * R1) + R2))
I = 2EMF / ((2 * R1) + R2))

I = 2EMF / ((2 * R1) + R2))

Part C

Calculate the current IB through the resistor of resistance R2 for circuit B.
Express the current in terms of EMF, R1, and R2.

This one is a little more complicated. You will have to apply Kirchhoff’s loop rule to each loop separately, and then apply Kirchhoff’s junction rule. If this isn’t immediately clear, take a look at the figure. There are basically two “loops” within circuit B. In either case, the bottom part of the circuit is the same – treat this portion of the circuit as segment “Z”. The first loop is the outermost pathway. This loop uses the upper wire at the top of the circuit – treat this upper wire as segment “X”. The second loop uses the lower wire at the top of the circuit. Treat that lower segment (at the top of the circuit) as segment “Y”. What you need to do is find the EMF for each loop separately:

For loop 1 (outermost) this gives you:

EMF = IX * R1 + IZ * R2

And or loop 2, you get:

EMF = IY * R1 + IZ * R2

Now, what you should notice is that the current in segment “Z” is just the sum of the current from segments “X” and “Y”:

IZ = IX + IY

Of course, ordinarily, the current in a wire is the same everywhere – but this circuit is actually several separate wires. The branches up top are basically the main wire (segment Z) cut into two different wires. Anyway, you need to solve for IB. Since the two separate wires join down at segment Z, you know that the current in segment Z is equal to IB. We need to solve by using relationship between IX and IY:

IX + IY = IZ

So:

IX + IY = IB

Now what?

Well, go back to the equations for loop 1 and loop 2, and solve for IX and IY:

EMF = IX * R1 + IZ * R2
EMF – (IZ * R2) = IX * R1
IX = EMF – (IZ * R2) / R1

And you’ll find that IY is the same thing:

IY = EMF – (IZ * R2) / R1

Substitute IB back in for IB:

IX = EMF – (IB * R2) / R1
IY = EMF – (IB * R2) / R1

Now we can really start solving. We know that IB is just the sum of IX and IY:

IB = IX + IY
IB = (EMF – (IB * R2) / R1) + (EMF – (IB * R2) / R1)
IB = [2*EMF - 2 * (IB * R2)] / R1
IB * R1 = [2*EMF - 2 * (IB * R2)]

Now divide by IB (this simplifies things a little):

R1 = (2*EMF / IB) – 2*R2
2*EMF/IB = R1 + 2*R2
2*EMF = IB * (R1 + 2*R2)
IB = 2*EMF / (R1 + 2*R2)

That was all just algebra, but it can be very tricky.

IB = 2*EMF / (R1 + 2*R2)

Part D

What is the power dissipated by the resistor of resistance R2 for circuit A, given that Ε = 10 V, R1 = 300 ohms, and R2 = 5000 ohms?
Calculate the power to two significant figures.

Remember that P = I^2 * R

Since we found the current for Circuit A in Part B, we can solve easily:

P = I^2 * R

I = 2EMF / ((2 * R1) + R2))

So:

P = [2EMF / ((2 * R1) + R2)]^2 * R2

Since this question is asking for the power dissipated in R2, we just substituted that for “R” in the P = I^2 * R formula.

So now solve:

P = [2 * EMF / ((2 * R1) + R2)]^2 * R2
P = [2 * 10 / ((2 * 300) + 5000)]^2 * 5000
P = [20 / 5600]^2 * 5000
P = 0.06377

Round to two significant figures:

P = 0.064 W

0.064 W

Part E

For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?

As long as the voltage and current are the same, the power will be the same if the resistors are equal. So the answer is, so long as:

R1 / R2 = 1

R1 / R2 = 1

Part F

Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B?

• R2 < R1
• R2 > R1
• R2 < 2R1
• R2 < 0.5R1

The best answer is “R2 > R1“, because as long as everything else is equal, more resistance = more power dissipation.

R2 > R1

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