Resistive Networks- AIPMT 2008 Questions

The following questions on resistive networks appeared in All India Pre-Medical/Pre-Dental Entrance Examination 2008 question paper:

(1) In the circuit shown, the current through the 4 Ω resistor is 1 amp when the points P and M are connected to a d.c. voltage source. The potential difference between the points M and N is

(1) 1.0 volt

(2) 0.5 volt

(2) 3.2 volt

(2) 1.5 volt

Since the current through the 4 Ω resistor is 1 ampere, the potential difference between the points P and M is 4 volt. The 1 Ω resistor and the two parallel resistors (each of value 0.5 Ω) make the total resistance in the branch PNM equal to 1.25 Ω. The potential drop across the 1 Ω resistor is therefore equal to 4×1/1.25 = 3.2 volt.

[The p.d. of 4 volt gets divided between 1 Ω and 0.25 Ω]

(2) A current of 3 amp flows through the 2 Ω resistor shown in the circuit. The power dissipated in the 5 Ω resistor is

(1) 2 watt

(2) 1 watt

(3) 5watt

(4) 4watt

The potential difference across the 2 Ω resistor is 2×3 = 6 volt.

The same p.d. appears across the branch containing 1 Ω and 5 Ω. Therefore, the current through 1 Ω and 5 Ω is 6 V/6 Ω = 1 A.

Therefore, the power dissipated in the 5 Ω resistor is I²R = 1²×5 = 5 watt.