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## Friday, April 11, 2008

### AP Physics B & C – Multiple Choice Questions (for practice) on Direct Current Circuits

The essential formulae to be remembered in connection with direct current circuits were given in the post daied 31st March 2008. A couple of free response questions were posted on 5th April 2008. Today we will discuss a few multiple choice questions in this section.

In the circuit shown, the battery has some internal resistance. Answer the following questions numbered (1) and (2) in respect of the circuit:

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 R1?

(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 R3. The current I flowing through R3 is therefore given by

I = 3 V/6 Ω = 0.5 A.

The current through R1 is the same as that through R3. 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/(R1+ R2+ R3+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 R1, R2, R3, R4 and R5 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 R1, R2, R3 and R4 make a balanced Wheatstone bridge since R1/R2 = R3/R4. No current can therefore pass through the resistor R5 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 (R4) is very nearly

(a) 0.567A

(b) 0.667 A

(c) 0.876 A

(d) 0.95 A

(e) 1 A

The resistor R5 can be ignored as we saw while answering question No.3. The current flowing through R4 is the current flowing through the lower parallel branch containing R3 and R4. 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 (I1) through one branch is given by

I1 = (Main current× Resistance of the other branch)/ Total resistance

In the present case current through the branch containing R4 = 2× (R1+R2)/ (R1+R2+ R3+R4) =

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