Two Kerala Engineering & Medical Entrance 2008 Questions on Resistive Networks

Questions involving resistive networks can be solved are often found in degree entrance examinations. Here is a reasonably good question which appeared in KEAM 2008 (Engineering) question paper:

In the circuit shown, if the resistance 5 Ω develops a heat of 42 J per second, the heat developed in 2 Ω must be about (in J s^–1)

(a) 25

(b) 20

(c) 30

(d) 35

(e) 40

The heat developed per second (power dissipation) in a resistance R is V²/R where V is the voltage across the resistance

[This expression has the forms I²R and VI where I is the current. Even though you can use these forms also to solve the problem, the form V²/R will be more convenient here]

Therefore we have

V²/5 = 42 watt

The power dissipated in the parallel branch containing the resistors 6 Ω and 9 Ω is V²/(6+9) = V²/15 and this must be 14 watt (since the denominator is 3 times)

The total power dissipated in the two parallel branches is thus 42+14 = 56 watt and the effective resistance of the two parallel branches is 5×15/(5+15) = 75/20 Ω.

If I is the current passing through the 2 Ω resistance, we have

I²×2 = W where W is the power dissipated in 2 Ω.

Since the same current passes through the effective resistance (75/20 Ω) of the parallel branches, we have

I²×75/20 = 56

From the above equations (on dividing),

40/75 = W/56 so that W = 40×56/75 = 30 watt nearly

The following MCQ which appeared in KEAM 2008 (Medical) question paper is simple:

In the Wheatstone’s network shown in the figure, the current I in the circuit is

(a) 1 A

(b) 2A

(c) 0.25 A

(d) 0.5 A

(e) 0.33 A

The bridge is balanced (since 2 Ω/4 Ω = 4 Ω/8 Ω). Therefore, the points B and D are equipotential points and there is no current through the diagonal 5 Ω resistor. The circuit thus reduces to 6 Ω and 12 Ω in parallel with the 2 V battery. The parallel combined value of 6 Ω and 12 Ω is 6×12/(6+12) = 4 Ω.

The current, I =2 V/4 Ω = 0.5 A.