Question

A p.d. of 6 V is applied to two resistors of 3 Ω and 6 Ω connected in parallel. Calculate
(a) the combined resistance
(b) the current flowing in the main circuit
(c) the current flowing in the 3 Ω resistor.

Answer 1

Answer:

a) 1/R= 1/2 +1/3 = 3/6 =1/2

R= 2 ohm

b) pd / R = 6/2=3ampere

c) pd/R =6/3=2 ampere

Answer 2

Explanation:

According to the question, the two resistors are connected in parallel and potential difference of 6 V is applied.

(a) The combined resistance, $\frac{1}{R}=\frac{1}{R 1}+\frac{1}{R 2}=\frac{1}{3}+\frac{1}{6}=\frac{3}{6}.$

(b) The current flowing in the main circuit, I.

From ohm’s law, V= IR

$I=\frac{V}{R}=\frac{6}{2}=3 \mathrm{A}.$

(c) The current flowing in the 3 ohm resistors, i.

$i=1 \times \frac{R 2}{R 1+R 2}=3 A \times \frac{6}{3+6}=3 \times \frac{6}{9}=2 \mathrm{A}$