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The following multiple choice questions appeared in Karnataka CET 2004 question paper:

**(1) In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the 2 Ω resistor is**

**(a) 0.6 A**

**(b) 1.2 A**

**(c) 0.9 A **

**(d) 1.5 A**

In steady state, the capacitor is fully charged and no current flows into its branch. So, you can ignore the branch containing the capacitor and the 4 Ω resistor. The circuit thus reduces to the 6 V cell in series with the 2.8 Ω resistor and the parallel combined value of 2 Ω and 3 Ω [ which is 2**×**3/(2+3) = (6/5)** **Ω = 1.2 Ω].

The current sent by the cell is 6 V/ (2.8 + 1.2) Ω = (6/4) A = 1.5 A.

This current gets divided between the 2Ω and 3Ω resistor paths. The current (*I*) through the 2Ω resistor is given by

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

[Remember the above relation for the branch current when current gets divided between two branches].

Therefore, we have

*I* = (1.5×3)/(2+3) = 0.9 A.

**(2) An unknown resistance R _{1} is connected in series with a resistance of 10**

**Ω. This combination is connected to one gap of a metre bridge while a resistance R**

_{2}is connected in the other gap. The balance point is at 50 cm. Now, when the 10**Ω resistor is removed and R**

_{1 }alone is connected, the balance point shifts to 40 cm. The value of R_{1}is (in ohms)**(a) 20**

**(b) 10**

**(c) 60**

**(d) 40**

The condition for the balance of a metre bridge (which is basically a Wheatstone bridge) is

P/Q = L/(100 – L) where P and Q are the resistances in the two gaps and the balancing length L cm is measured from the side of P.

In the above problem we have

(R_{1} + 10) /R_{2} = 1 since the balance point is at the middle of the wire when one gap contains the series combination of R_{1} and 10 Ω and the other gap contains R_{2}.

With R_{1 }alone in the gap (on removing the 10 Ω resistance), the balance condition is

R_{1}/R_{2 }= 40/60

From the above two equations (on dividing one by the other), we have

(R_{1} + 10) /R_{1} = 60/40, from which R_{1} = 20 Ω.