Question

Four resistors, a voltmeter and a battery are connected in a circuit as shown below.

(a) What is the net resistance in the circuit? 

(b) How much potential difference will the voltmeter connected across the resistor R₄ measure?

OR 

What is the power dissipated by the resistor R₁? 

(c) If R₃ is removed, will the net current in the circuit increase or decrease or remain the same? Justify your answer.

Answer 1

(a) The net resistance is: R₁ + (1/R₂ + 1/R₃) + R₄ 

= 15 + 10 + 15 

R = 40 Ω 

(b) Voltage drop across R₄ = Net current x R₄ 

Net current = V/R 

= 20/40 

= 0.5 A  

Voltage drop across R₄ = 0.5 x 15 

= 7.5 V 

OR 

Power dissipated by the resistor R₁ is given by: 

P = I²R₁ 

I = V/R = 20/40 

I = 0.5 A  

Therefore, Power = (0.5)² x 15 = 3. 75 W 

(c) Net current will decrease because R₃ is connected in parallel and removing it will increase the net resistance in the circuit thereby reducing the net current.