Solved Sample Paper 2 — Question 13

Given, bulbs of 40 Ω and 70 Ω are connected in series.

Let their equivalent resistance be R_s then

R_s = 40 + 70 = 110 Ω

Hence, resistance in the circuit = 110 Ω

(b) According to Ohm's law: V = IR

Substituting the values in the formula above, we get

220 = I x 110

I = $\dfrac{220}{110}$ = 2 A

Hence, current in the circuit = 2 A

(c) Voltage drop across the 40 Ω bulb (B₁) = ?

I = 2A

Applying Ohm's Law

V = 2 x 40 = 80 V

Voltage drop across the 70 Ω bulb (B₂) = ?

I = 2A

Applying Ohm's Law

V = 2 x 70 = 140 V

Hence, voltage drop across B₁ is 80 V and across B₂ is 140 V.