Two AIEEE 2005 Questions on Direct Current Circuits

(1) In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

(a) 1000Ω (b) 500Ω (c) 100Ω (d) 200Ω

As there is no current through the galvanometer, the voltage drop across R must be 2V ( so as to balance the emf 2V of the battery B). The voltage drop (produced by the current sent by battery A) across 500Ω is therefore 10V. The same current flows through R and hence R must be 100Ω.

(2) An energy source will supply a constant current in to the load if its internal resistance is

(a) non zero but less than the resistance of the load

(b) zero

(c) very large compared to the load resistance

(d) equal to the resistance of the load

The correct option is (c) since the current will not change with the load resistance if the total resistance of the circuit is almost equal to the source resistance. Ideally, a constant current generator should have infinite resistance.

[You may note also that an ideal constant voltage generator should have zero resistance].

Direct Current Circuits

The following simple question appeared in Kerala Medical Entrance 2006 test paper:
Four cells each of emf 2V and internal resistance 1Ω are connected in parallel to a load resistor of 2Ω. Then the current through the load resistor is
(a) 2A (b) 1.5A (c) 1A (d) 0.888A (e) 0.75A
Since the cells are connected in parallel, the internal resistance of the combination is 1/4=0.25Ω. If ‘R’ is the load resistance and ‘r’ is the internal resistance, the current through load resistance= V/(R+r) = 2/(2+0.25) = 0.888A. So (d) is the correct option.
Now consider the following M.C.Q. which appeared in the question paper of Karnataka C.E.T.2003:
Two wires of the same dimensions but resistivities ρ₁ and ρ₂ are connected in series. The equivalent resistivity of the combination is
(a) 2(ρ₁ + ρ₂) (b) √(ρ₁ρ₂) (c) (ρ₁ + ρ₂)/2 (d) ρ₁ + ρ₂
If R₁ and R₂ are the resistances of the wires, R₁ = ρ₁L/A and R = ρ₂L/A where L and A are the length and the cross section area respectively. When the wires are connected in series the total length is 2L and the total resistance is R₁ + R₂. Hence the resistivity of the combination is given by ρ = (R₁ + R₂)A/2L = (R₁A/2L) + (R₂A/2L) = ρ₁/2 + ρ₂/2. The correct option therefore is (c).
Suppose we make this question a bit more complicated by making the lengths of the two wires different (say, L₁ and L₂). Then, ρ = (R₁ + R₂)A/(L₁ + L₂) = [(ρ₁L₁/A) + (ρ₂ L₂/A)] A/(L₁ + L₂) = (ρ₁L₁ + ρ₂L₂)/(L₁ + L₂).
Now, consider the following question:
The heat produced per second in a 2Ω resistor on connecting it across a battery is the same as the heat produced per second in an 8Ω resistor on connecting it across the same battery after disconnecting the 2Ω resistor. The internal resistance of the battery is
(a) 4Ω (b) 5Ω (c) 6Ω (d) √10 Ω (e) 3.3Ω
It is convenient to remember the internal resistance as √(R1R2) in situations like this. The answer is √(2×8) = 4 Ω. You can work it out from ‘first principles’ like this:
The currents through the resistors in the two cases are V/(2 + r) and V/(8 + r) with usual notations. Equating the powers (I2R values), we have, [V/(2 + r)]2×2 = [V/(8 + r)]2×8, from which r = 4 Ω.
Let us consider the following question also taken from Karnataka C.E.T.2003 question paper:
If a 30V, 90W bulb is to be worked on 120V line, the resistance to be connected in series with the bulb is
(a) 40 Ω (b) 30 Ω (c) 20 Ω (d) 10 Ω
30V, 90W bulb means the power consumed by the bulb on connecting it across a 30volt supply is 90 watt. The current drawn by this bulb (while incandescent) is 90watt/30 volt = 3 ampere and the resistance of this bulb is 30V/3A = 10 Ω. The problem is basically to find out the resistance required to limit the current to 3A. If R is the resistance required in series with the bulb, we have, 120volt/(10 + R)ohm = 3 ampere. Therefore, R = 30 Ω.