The essential formulae to be remembered in connection with direct current circuits were given in the post daied ^{st} March 2008^{th} April 2008

**Question No.(1):**

The switch S has been closed for a long time and the capacitor C has been fully charged. If the voltage across the capacitor is found to be 3 V, what is the current flowing through the resistor R_{1}?

(a) zero

(b) 0.2 A

(c) 0.5 A

(d) 0.6 A

(e) 6/11 A

Since the capacitor is fully charged, the voltage across the capacitor is equal to the steady voltage across the 6 Ω resistor R_{3}. The current *I *flowing through R_{3} is therefore given by

*I *= 3 V/6 Ω = 0.5 A.

The current through R_{1} is the same as that through R_{3}. So the correct option is 0.5 A.

**Question No.(2):**

What is the internal resistance of the battery?

(a) 0.2 Ω

(b) 0.5 Ω

(c) 1 Ω

(d) 18 Ω

(e) 6/11 A

The current *I *delivered by the battery can be written as

*I* = E/(*R*_{1}+* R*_{2}+* R*_{3}+*r*) where *E** *is the emf and *r* is the internal resistance of the battery. Since the current *I* is 0.5 ampere, we have

0.5 A = 9V/(5+6+6+*r*)Ω so that *r* = 1 Ω.

*Question numbers (3), (4) and (5) given below relate to the adjoining circuit in which a current of 2A enters the network of resistors R*

_{1}, R_{2}, R_{3}, R_{4}and R_{5}through the junction P and leaves through the junction Q.**Question No.(3):**

The effective resistance between the junctions P and Q is

(a) 1 Ω

(b) 2 Ω

(c) 3 Ω

(d) 4 Ω

(e) 5 A

The resistors R_{1}, R_{2}, R_{3} and R_{4} make a balanced Wheatstone bridge since R_{1}/R_{2} = R_{3}/R_{4}. No current can therefore pass through the resistor R_{5} so that it can be ignored. The circuit therefore reduces to 3Ω (which is the series combined value of 1Ω and 2Ω) in parallel with 6Ω (which is the series combined value of 2Ω and 4Ω). Therefore, the effective resistance between the junctions P and Q is (3×6)/(3+6) = 2 Ω.

**Question No.(4):**

The potential difference between the junctions P and Q is

(a) 2 V

(b) 4 V

(c) 6 V

(d) 8 V

(e) 10 V

Since the effective resistance between the junctions P and Q is 2 Ω (as shown above) and the current flowing into the network is 2 A, the potential difference (IR drop) between the junctions P and Q is 2×2 = 4 V.

**Question No.(5):**

The current flowing through the 4 Ω resistor (R_{4}) is very nearly

(a) 0.567A

(b) 0.667 A

(c) 0.876 A

(d) 0.95 A

(e) 1 A

The resistor R_{5} can be ignored as we saw while answering question No.3. The current flowing through R_{4} is the current flowing through the lower parallel branch containing R_{3} and R_{4}. The lower parallel branch has total resistance 6 Ω and the upper branch has total resistance 3 Ω. The current being inversely proportional to the resistance of the branch, the main current of 2A must get divided in the ratio 3:6 (= 1:2) between the lower and upper branches. The current through the lower branch must therefore be 2×(1/3) =2/3 A which is very nearly equal to 0.667 A.

[When current gets divided between two parallel branches, the current (*I*_{1}) through one branch is given by

*I*_{1} = (Main current× Resistance of the other branch)/ Total resistance

In the present case current through the branch containing R_{4 }= 2× (R_{1}+R_{2})/ (R_{1}+R_{2}+ R_{3}+R_{4}) =

2× (1+2)/(1+2+2+4) = 2/3 = 0.667 A, very nearly].

You will find similar multiple choice questions (MCQ) with solution from different branches of physics at **physicsplus.blogspot.com**

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