The diagram shows an electric circuit in which the resistance of the external resistor is 2R and the internal resistance of the source is R.

Question 7

The diagram shows an electric circuit in which the resistance of the external resistor is 2R and the internal resistance of the source is R.

What is the ratio          power in external resistance / power in internal resistor ?

A ¼                 B ½                 C 2                  D 4

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q33

Solution:

Answer: C.

P = I²R

where I is the current flowing through resistor R

In the circuit, the same current I flows as the resistor may be considered to be connected in series to the battery.

P_ext = I²(2R) = 2 I²R

P_int = I²R

So, ratio = P_ext / P_int = 2 I²R / I²R = 2

A uniform solid block has weight 500 N, width 0.4 m and height 0.6 m. The block rests on the edge of a step of depth 0.8 m, as shown.

Question 8

A uniform solid block has weight 500 N, width 0.4 m and height 0.6 m. The block rests on the edge of a step of depth 0.8 m, as shown.

The block is knocked over the edge of the step and rotates through 90° before coming to rest with the 0.6 m edge horizontal.

What is the change in gravitational potential energy of the block?

A 300 J                        B 400 J                        C 450 J                       D 550 J

Reference: Past Exam Paper – June 2015 Paper 12 Q15

Solution:

Answer: C.

Since the block is said to be uniform, the centre of mass can be considered to be at the centre.

At the top of the step,

Height of position of centre of mass above ground = 0.8 + (0.6/2) = 0.8 + 0.3 = 1.1 m

At the bottom of the step,

Height of position of centre of mass above ground = 0.4/2 = 0.2 m

Change in GPE = mgΔh = 500 × (1.1 – 0.2) = 450 J

Three cells with e.m.f.s V1, V2 and V3, have negligible internal resistance. These cells are connected to three resistors with resistances R1, R2 and R3, as shown.

Question 6

Three cells with e.m.f.s V1, V2 and V3, have negligible internal resistance. These cells are

connected to three resistors with resistances R1, R2 and R3, as shown.

The current in the circuit is I.

Which equation is correct?

A V1 + V2 + V3 = I (R1 + R2 + R3)

B V1 + V2 – V3 = I (R1 + R2 + R3)

C V1 – V2 + V3 = I (R1 + R2 + R3)

D V1 – V2 – V3 = I (R1 + R2 + R3)

Reference: Past Exam Paper – March 2016 Paper 12 Q34

Solution:

Answer: D.

The sum of p.d. across the resistors in the circuit is equal to the overall e.m.f. in the circuit.

The current I is shown to flow from the positive terminal of cell V₁. So, cell V₁ is taken as the reference.

For cells to be connected correctly, the positive terminal of one cell should be connected to the negative terminal of the other. But the negative terminal of V₃ is connected to the negative terminal of V₁ and also, the positive terminal of V₂ is connected to the positive terminal of V₁. Thus, both V₂ and V₃ are wrongly connected to V₁ and hence, the two cells are reducing.

 

Overall p.d. = V₁ – V₂ – V₃ = I(R₁ + R₂ + R₃)