Resistors in series and parallel; power

Explanation of question 11

In Fig. 3.4, the p.d. between A and B, and the total power dissipated, will be respectively

  1. 24V, 240mW
  2. 12V, l20mW
  3. 6V, 60mW
  4. 6V, 30mW

The answer is :- 2

All current from the 12 V battery flows through the lower 1.2 k resistor. No current flows through upper 1.2 k resistor, so there is no p.d. across the upper 1.2 k resistor.

Potential at A is the same as the potential of the positive terminal of the battery,

i.e. 12 V, therefore p.d. between A and B = 12 V.

The only power dissipated will be the lower 1.2 k resistor and will be:

P=V2 / R =

12 2
1.2x103

= 144 / 1.2 mW = 120 mW

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