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 

= 0.2 A  

Voltage drop across R₄ = 0.2 x 15 

= 3 V 

OR 

Power dissipated by the resistor R1 is given by: 

P = I²R₁ 

I = V/R = 20/40 

I = 0.2 A  

Therefore, Power = (0.2)² x 15 = 0.6 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.