**Batteries in Series or Parallel**

Part A = counterclockwise

Part B = I = 2EMF / ((2 * R_{1}) + R_{2})

Part C = I_{B} = 2*EMF / (R_{1} + 2*R_{2})

Part D = 0.064 W

Part E = R_{1} / R_{2} = 1

Part F = R_{2} > R_{1}

**Solution Below:**

_{1}each. Circuit A has the batteries connected in series with a resistor of resistance R

_{2}, 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

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

_{2}in circuit A?

Express the current in terms of EMF, R

_{1}, and R

_{2}.

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 * R_{1}) + R_{2}))

I = 2EMF / ((2 * R_{1}) + R_{2}))

I = 2EMF / ((2 * R_{1}) + R_{2}))

Part C

_{B}through the resistor of resistance R

_{2}for circuit B.

Express the current in terms of EMF, R

_{1}, and R

_{2}.

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 = I_{X} * R_{1} + I_{Z} * R_{2}

And or loop 2, you get:

EMF = I_{Y} * R_{1} + I_{Z} * R_{2}

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

I_{Z} = I_{X} + I_{Y}

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 I_{B}. Since the two separate wires join down at segment Z, you know that the current in segment Z is equal to I_{B}. We need to solve by using relationship between I_{X} and I_{Y}:

I_{X} + I_{Y} = I_{Z}

So:

I_{X} + I_{Y} = I_{B}

Now what?

Well, go back to the equations for loop 1 and loop 2, and solve for I_{X} and I_{Y}:

EMF = I_{X} * R_{1} + I_{Z} * R_{2}

EMF – (I_{Z} * R_{2}) = I_{X} * R_{1}

I_{X} = EMF – (I_{Z} * R_{2}) / R_{1}

And you’ll find that I_{Y} is the same thing:

I_{Y} = EMF – (I_{Z} * R_{2}) / R_{1}

Substitute I_{B} back in for I_{B}:

I_{X} = EMF – (I_{B} * R_{2}) / R_{1}

I_{Y} = EMF – (I_{B} * R_{2}) / R_{1}

Now we can really start solving. We know that I_{B} is just the sum of I_{X} and I_{Y}:

I_{B} = I_{X} + I_{Y}

I_{B} = (EMF – (I_{B} * R_{2}) / R_{1}) + (EMF – (I_{B} * R_{2}) / R_{1})

I_{B} = [2*EMF - 2 * (I_{B} * R_{2})] / R_{1}

I_{B} * R_{1} = [2*EMF - 2 * (I_{B} * R_{2})]

Now divide by I_{B} (this simplifies things a little):

R_{1} = (2*EMF / I_{B}) – 2*R_{2}

2*EMF/I_{B} = R_{1} + 2*R_{2}

2*EMF = I_{B} * (R_{1} + 2*R_{2})

I_{B} = 2*EMF / (R_{1} + 2*R_{2})

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

I_{B} = 2*EMF / (R_{1} + 2*R_{2})

Part D

_{2}for circuit A, given that Ε = 10 V, R

_{1}= 300 ohms, and R

_{2}= 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 * R_{1}) + R_{2}))

So:

P = [2EMF / ((2 * R_{1}) + R_{2})]^2 * R_{2}

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

So now solve:

P = [2 * EMF / ((2 * R_{1}) + R2)]^2 * R_{2}

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

_{1}and R

_{2}would power dissipated by the resistor of resistance R

_{2}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:

R_{1} / R_{2} = 1

R_{1} / R_{2} = 1

Part F

_{2}in circuit A be bigger than that of circuit B?

- R
_{2}< R_{1} **R**_{2}> R_{1}- R
_{2}< 2R_{1} - R
_{2}< 0.5R_{1}

The best answer is “R_{2} > R_{1}“, because as long as everything else is equal, more resistance = more power dissipation.

R_{2} > R_{1}