# IBDP Physics 5.2 – Heating effect of electric currents: IB Style Question Bank SL Paper 1

### Question

Two conductors S and T have the V/I characteristic graphs shown below. When the conductors are placed in the circuit below, the reading of the ammeter is 6.0 A. What is the emf of the cell?

A 4.0 V

B 5.0 V

C 8.0 V

D 13 V

Ans: A

The  equation  $$V=iR$$ is the defining equation for resistance, and it applies to all conducting devices, whether they obey Ohm’s law or not. If we measure the potential difference $$V$$ across, and the current $$i$$ through, any device, even a pn junction diode, we can find its resistance at that value of $$V$$  as $$R=\frac{V}{i}$$

$$\frac{1}{R_T}=\frac{6-0}{8-0}=\frac{3}{4}$$

From the graph , V and I is not defined for V equal 8 0r 13 hence these can not be correct options.

If emf = 5 V

Then from graph we have

$$\frac{1}{R_S} =\frac{6}{5}$$
hence
$$\frac{1}{R} =\frac{1}{R_S} +\frac{1}{R_T} =\frac{6}{5}+\frac{6}{8} =1.95$$
$$i=v \times \frac{1}{R} =5\times 1.95 =9.75 \neq 6$$
but current is 6 A hence this can not be correct option

Now verify for emf = $$4v$$

$$\frac{1}{R_S} =\frac{3}{4}$$

$$\frac{1}{R_T} =\frac{6}{8}$$

$$\frac{1}{R} =\frac{1}{R_S} +\frac{1}{R_T} =\frac{3}{4}+\frac{6}{8} =\frac{3}{2}$$

$$i=v \times \frac{1}{R} =4\times \frac{3}{2} =6$$

which is true as given current is 6 A. Hence correct option is A

### Question

An electric motor raises an object of weight 500 N through a vertical distance of 3.0 m in 1.5 s. The current in the electric motor is 10 A at a potential difference of 200 V. What is the efficiency of the electric motor?

A. 17 %

B. 38 %

C. 50 %

D. 75 %

Ans: C

Electric energy supplied by Motor = $$Pt=VIt$$

Given V= 200v , I = 10 A and t = 1.5 s, weight = 500 N ,vertical distance =3m

Electric energy $$E = VIt = 200 \times 10\times 1.5 =3000 \; J$$

Now Work Output = $$mgh = 500 \times 3 = 1500 \; J) \(efficiency = \frac{Output }{Input}\times 100= \frac{1500}{3000}\times 100 = 50 \;%$$

### Question

Three resistors are connected as shown. What is the value of the total resistance between X and Y? A.     1.5 Ω

B.     1.9 Ω

C.     6.0 Ω

D.     8.0 Ω

### Markscheme

A $$3\Omega$$ resistance are in series , equivalent of which, is in parallel to $$2\Omega$$

$$R_1= 3+3 = 6 \Omega$$
$$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{2}=\frac{1}{6}+\frac{1}{2} =\frac{4}{6}$$
$$\therefore R= \frac{6}{4}=1.5 \Omega$$

### Question

Kirchhoff’s laws are applied to the circuit shown. What is the equation for the dotted loop?

A. 0 = 3I2 + 4I3

B. 0 = 4I3 − 3I2

C. 6 = 2I1 + 3I2 + 4I3

D. 6 = 3I2 + 4I3

### Markscheme

B Along the loop $$IR$$ is taken positive and opposite to it negative. Also emf in this loop is zero as there is no external cell

Hence

$$3\times I_2 -4\times I_3 = 0$$
$$or$$
$$4\times I_3 -3\times I_2 =0$$