Question

A p.d. of 4 V is applied to two resistors of 6 Ω and 2 Ω connected in series. Calculate
(a) the combined resistance
(b) the current flowing
(c) the p.d. across the 6 Ω resistor

Answer 1

a) The combined resistance is 8 Ω.

b) The current flowing through circuit is 0.5 Ampere.

c) The P.D. across the 6 Ω resistor is 1.5 watts .

Explanation:

Given:

A p.d. of 4 V is applied to two resistors of 6 Ω and 2 Ω connected in series.

a)The combined resistance (Re) is given by

Since , the resistance is connected in series then the equivalent resistance is sum of all resistance.

Re = 6 Ω + 2 Ω

Re = 8 Ω

b) Since,

$v = i \times r$

$4 = i \times 8$

$i = \frac{4}{8}$

$i = 0.5$

The current flowing through circuit is 0.5 Ampere.

c) The power is given by,

$p = {i}^{2} \times r$

For p.d. across 6 Ω is given by

$p = {0.5}^{2} \times 6$

$p = 1.5w$

The P.D. across the 6 Ω resistor is 1.5 watts .

Answer 2

Combined Resistance is 2 $\Omega$

Explanation:

• The resistors of $6 \Omega$  and $3 \Omega$  are connected in parallel.  

Therefore, their combined resistance can be calculated as

$\frac {1}{R} = \frac {1}{R_1} + \frac {1}{R_2}$

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

R = $2 \Omega$

(b) The p.d. across the combined resistance, V= IR

Here I = 6A and combined resistance = $2 \Omega$

So, V = $6 \times 2$

= 12 V

(c) In parallel connection potential difference remains constant, so Potential difference across $3 \Omega$ resistor = 12 V

(d) The current in the $3 \Omega$ resistor

$I = \frac {V}{R_1}$

$I = \frac {12}{3}$

= 4 A

(e) Current flowing through the $6 \Omega$ resistor = $I = \frac {V}{R_2}$

$I = \frac {12}{6}$

= 2 A