# IBDP Physics 5.3 – Electric cells: IB Style Question Bank HL Paper 1

### Question

For a real cell in a circuit, the terminal potential difference is at its closest to the emf when

A the internal resistance is much smaller than the load resistance.

B a large current flows in the circuit.

C the cell is not completely discharged.

D the cell is being recharged.

Ans: A

Potential difference and emf are two different quantities whose magnitudes may be equal in certain conditions. The emf is the work done per unit charge by the battery force Fb which is nonelectrostatic in nature. The potential difference originates from the electrostatic field created by the charges accumulated on the terminals of the battery.

The terminal potential difference of a cell is equal to the emf of the cell when:

The cell is an open circuit.
The internal resistance of the cell is zero.
The load resistance much greater than internal resistance

### Question

When an electric cell of negligible internal resistance is connected to a resistor of resistance 4R, the power dissipated in the resistor is P.

What is the power dissipated in a resistor of resistance value R when it is connected to the same cell?

A. $$\frac{P}{4}$$

B. P

C. 4P

D. 16P

### Markscheme

C

Power dissipated $$P$$ is $$Vi$$
Given For resistance $$= 4R$$ , Power  $$= P$$
Hence
$$P = Vi = V\times\frac{V}{4R}=\frac{V^2}{4R}$$
now for Resistance $$R$$
$$P{}’=Vi{}’ = V\times\frac{V}{R}=\frac{V^2}{R}$$
$$\frac{P{}’}{P}=\frac{\frac{V^2}{R}}{\frac{V^2}{4R}}=4$$
$$\therefore P{}’=4 P$$

### Question

A 12V battery has an internal resistance of 2.0Ω. A load of variable resistance is connected across the battery and adjusted to have resistance equal to that of the internal resistance of the battery. Which statement is correct for this circuit?

A. The current in the battery is 6A.
B. The potential difference across the load is 12V.
C. The power dissipated in the battery is 18W.
D. The resistance in the circuit is 1.0Ω.

### Markscheme

C

Total Resistance in circuit $$R= 2+2 =4 \Omega$$

current in circuit
$$i=\frac{E}{R}=\frac{12}{4} = 3\; A$$
$$=i\times R_{load}= 3\times 2 = 6\; V$$
Power dissipated in battery $$= i^2 r=3^2\times 2= 18 \; W$$

### Question

A circuit is formed by connecting a resistor between the terminals of a battery of electromotive force (emf) 6 V. The battery has internal resistance. Which statement is correct when 1 C of charge flows around the complete circuit?

A. 6 V is the potential difference across the resistor.

B. 6 J of thermal energy is dissipated in the battery.

C. 6 J of chemical energy is transformed in the battery.

D. 6 J of thermal energy is dissipated in the resistor.

### Markscheme

C

Given $$q =1$$
R = Unknown
Hence potential across resistor can not be calculated
Work Done by Battery due to chemical c is given by
$$E = \frac{W}{q}$$ , By definition of battery emf
$$or$$

$$W = E\times q = 6\times 1= 6 J$$

If a charge q is taken from the terminal B to the terminal A, the work done by the battery force Fb is W = Fb d where d is the distance between A and B. The work done by the battery force per unit charge is

$$E = \frac{W}{q}=\frac{F_b\times d}{q}$$

This quantity is called the emf of the battery. The full form of emf is electromotive force.