Q: A single cell of e.m.f. 1.5 V is connected across a 0.30 $$\Omega$$ resistor. The current in the circuit is 2.5 A. Calculate the terminal potential difference (p.d.) across the resistor.

$$0.30\ \Omega \times 2.5\ \mathrm{A} = 0.75\ \mathrm{V}$$, but the correct terminal p.d. is $$0.75\ \mathrm{V}$$

Q: The power dissipated in the external resistor is a maximum when its resistance $$R is $$equal to the internal resistance $$r of $$the cell. If $$E = 1.5\ \mathrm{V}$$ and $$r = 0.30\ \Omega$$, calculate the power dissipated when $$R = r.$$

$$P = \frac{E^2$}{4r} = \frac{(1.5)^2$}{4 \times 0.30} = 1.875\ \mathrm{W}$$

Question 1

A single cell of e.m.f. 1.5 V is connected across a 0.30 $$\Omega$$ resistor. The current in the circuit is 2.5 A. Calculate the terminal potential difference (p.d.) across the resistor.

Accepted Answer

$$0.30\ \Omega 	imes 2.5\ \mathrm{A} = 0.75\ \mathrm{V}$$, but the correct terminal p.d. is $$0.75\ \mathrm{V}$$

Question 2

Why is the terminal p.d. across the resistor not equal to the e.m.f. of the cell?

Accepted Answer

Because some energy is lost due to the internal resistance of the cell.

Question 3

Given a cell of e.m.f. $$E = 1.5\ \mathrm{V}$$, a resistor $$R = 0.30\ \Omega$$, and a current $$I = 2.5\ \mathrm{A}$$, show that the internal resistance $$r of $$the cell is $$0.30\ \Omega.$$

Accepted Answer

Using $$E = I(R + r)$$, $$r = \frac{E}{I} - R = \frac{1.5}{2.5} - 0.30 = 0.30\ \Omega.$$

Question 4

The power dissipated in the external resistor is a maximum when its resistance $$R is $$equal to the internal resistance $$r of $$the cell. If $$E = 1.5\ \mathrm{V}$$ and $$r = 0.30\ \Omega$$, calculate the power dissipated when $$R = r.$$

Accepted Answer

$$P = \frac{E^2$}{4r} = \frac{(1.5)^2$}{4 	imes 0.30} = 1.875\ \mathrm{W}$$

Question 5

Show that the power dissipated in the external resistor is less than the maximum when $$R = 0.50\ \Omega$$ and $$R = 0.20\ \Omega$$ for a cell with $$E = 1.5\ \mathrm{V}$$ and $$r = 0.30\ \Omega.$$

Accepted Answer

For $$R = 0.50\ \Omega$$, $$P = \frac{E^2$ R}{(R + r)^2$} = \frac{(1.5)^2$ 	imes 0.50}{(0.50 + 0.30)^2$} = 1.758\ \mathrm{W}$$; for $$R = 0.20\ \Omega$$, $$P = 0.9375\ \mathrm{W}. $$Both are less than $$1.875\ \mathrm{W}.$$

Question 6

A resistor of $$6.0\ \Omega$$ and a second resistor of $$3.0\ \Omega$$ are connected in parallel across a battery of e.m.f. $$4.5\ \mathrm{V}$$ and internal resistance $$0.50\ \Omega. $$What is the current in the battery?

Accepted Answer

$$I = \frac{4.5}{2.0 + 0.5} = 1.8\ \mathrm{A} ($$where $$R_{eq} = 2.0\ \Omega)$$

Question 7

In the potential divider circuit shown, what happens to the resistance of the thermistor and $$V_{out}$$ when the temperature decreases?

Accepted Answer

The resistance of the thermistor increases and $$V_{out}$$ increases.

Question 8

Explain what is meant by the internal resistance of a cell.

Accepted Answer

It is the resistance inside the cell that causes energy loss as heat when current flows.

Question 9

A cell is connected in series with a $$2.00\ \Omega$$ resistor and the current is $$0.625\ \mathrm{A}. If a $$second $$2.00\ \Omega$$ resistor is added in series, the current falls to $$0.341\ \mathrm{A}. $$Calculate the internal resistance of the cell.

Accepted Answer

$$r = 1.00\ \Omega$$

Question 10

Given the above scenario, calculate the e.m.f. of the cell.

Accepted Answer

$$E = 2.25\ \mathrm{V}$$

Question 11

A car battery needs to supply a current of $$200\ \mathrm{A} to $$turn over the starter motor. Explain why a battery made of a series of cells with $$r = 1.00\ \Omega$$ would not be suitable.

Accepted Answer

Because the internal resistance is too high, causing a large voltage drop and insufficient current.

Question 12

Which equation correctly relates the terminal p.d. $$V of a $$cell to its e.m.f. $$E$$, current $$I$$, and internal resistance $$r$$?

Accepted Answer

$$V = E - Ir$$

Question 13

If a thermistor is used in a potential divider circuit to control a fan, what will happen to the fan as the temperature increases?

Accepted Answer

The resistance of the thermistor decreases, $$V_{out}$$ increases, and the fan speeds up.

Question 14

A battery with e.m.f. $$E$$ and internal resistance $$r is $$connected to a load resistor $$R. $$Which expression gives the current $$I in $$the circuit?

Accepted Answer

$$I = \frac{E}{R + r}$$

Question 15

A $$4.5\ \mathrm{V}$$ battery with internal resistance $$0.50\ \Omega is $$connected to two resistors, $$6.0\ \Omega$$ and $$3.0\ \Omega$$, in parallel. What is the equivalent resistance of the parallel combination?

Accepted Answer

$$R_{eq} = 2.0\ \Omega$$

Internal Resistance, Circuits, and Potential Dividers SW 1

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