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

(a) The combined resistance is 2 Ω.

(b) The current flowing in the main circuit is 3 ampere.

(c) The current flowing in the 3 Ω resistor is 2 ampere.

Explanation:

Given:

A p.d. of 6 V is applied to two resistors of 3 Ω and 6 Ω connected in parallel.

(a) For parallel combination, the combined resistance is given by

$r = \frac{(r1 \times r2)}{r1 + r2}$

$r = \frac{3 \times 6}{3 + 6}$

$r = 2 Ω$

The combined resistance is 2 Ω.

(b) The current through the main circuit is given by

$v = i \times r$

$6 = i \times 2$

$i = 3$

The current flowing in the main circuit is 3 ampere.

(c)The current flowing in the 3 Ω resistor is given by

$i1 = i \times \frac{r2}{r1 + r2}$

$i1 = 3 \times \frac{6}{9}$

$i1 = 2$

The current flowing in the 3 Ω resistor is 2 ampere.

Answer 2

Combined Resistance is 2 $\Omega$

Explanation:

Given, $R_1 = 3 \Omega$

$R_2 = 6 \Omega$

V = 6 V

i) $\frac {1}{R_P} = \frac {1}{R_1} + \frac {1}{R_2}$

= $\frac {1}{3} + \frac {1}{6}$

So,  

$\frac {1}{R_P} = \frac {2+1}{6}$

= $\frac {3}{6} = \frac{1}{2}$

$R_P = 2 \Omega$

ii) $V = IR_P$

I = $\frac {V}{R_P}$

I = $\frac {6}{2}$

I = 3 A

iii) V = 6V

R = $3 \Omega$

I = $\frac {V}{R}$

= $\frac {6}{3}$

= 2 A