Question

Three resistors of resistances 5Ω, 10Ωand 15Ωconnected in series to a battery of 30V and to a key. Find the(i) current flowing in the circuit (ii)potential difference across each resistor. ​

Answer 1

Answer:

(i) 1 A

(ii)  5 V, 10 V and 15 V

Explanation:

Given, R1 = 5 ohm , R2 = 10 ohm, R3 = 15 ohm;

V = 30 V (R1 R2 and R3 are resistances and V is the voltage of the battery)

Total Resistance (R) = R1 + R2 + R3 (Since the resistors are in series)

= 5+ 10+15

= 30 ohm

(i) Current flowing through the external circuit(I) = V/R (Ohm's Law)

  = 30/30

  = 1 A

(ii) Since the resistors are in series the current flowing through them is same but their potential differences are different,

According to Ohm's Law,

Potential Difference across R1 = I×R1

= 1 × 5

= 5V

Potential Difference across R2 = I×R2

= 1×10

= 10V

Potential Difference across R3 = I×R3

= 1×15

= 15V