## Friday, March 16, 2007

### Questions (MCQ) on Direct Current Circuits

Here is a question which you can work out using Ohm’s law only:
Two constantan wires P and Q have their lengths in the ratio 1:2 and radii in the ratio 2:1. They are connected in series and potentials 2V and 20V are applied at the free ends of P and Q respectively. The potential at the junction of the wires is
(a) 2V (b) 4V (c) 9V (d) 12V (e) 16V
The resistances of P and Q are in the ratio 1:8 since the length of Q is twice that of P and the area of cross section of Q is a quarter if that of P. [The resistance is given by R = ρL/A where ρ is the resistivity, L is the length and A is the area of cross section].
The potential drops across P and Q (when current flows in them) are in the ratio 1:8. The potential difference applied across the series combination of P and Q is (20–2) = 18V. Therefore, the potential drop across Q = 18×8/(1+8) = 16V.
The potential at the junction of P and Q is (20V– 16V) = 4V.
Now, consider the following MCQ:
In the circuit shown, the power dissipated in the 2Ω resistor is 9W. What is the power dissipated in the 4Ω resistor?
(a) 18W (b) 12W (c) 9W (d) 3W (b) 2W
If the current through the 2Ω resistor is ‘I’, the current through the 4Ω resistor is I/3 since the total resistance (9Ω) in the 4Ω resistor branch is 3 times the total resistance (3Ω) in the 2Ω resistor branch. The expressions for power dissipation in 2Ω and 4Ω are respectvely
P = I2 ×2 and
P' = (I/3)2 ×4 so that P'/P = 2/9 from which P'= P×(2/9) = 9×2/9 = 2W.

#### 1 comment:

1. Sir,
Thanks for the extraordinary effort taken by you to help students through the blog.
The problems posted are very conceptual and there solutions too are crystal clear and self explanatory.

plz continue with this Great work.