Current Electricity — Question 27

(a) Given,

e.m.f. = 16 V

internal resistance r = 2 Ω

current through battery = ?

If R_p is the equivalent resistance of resistors 3 Ω and 6 Ω connected in parallel, then

$\dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2 + 1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{3}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{2} \\[0.5em] \Rightarrow R_p = 2 Ω$

From relation,

ε = I (R + r)

Substituting the value in the formula above we get,

16 = I(2 + 2)
⇒ 16 = I x 4 ⇒ I = 16 / 4 = 4 A

Hence, current through the battery = 4 A

(b) Potential difference between the terminals of the battery = ?

Using Ohm's law

V = IR

R = 2 Ω

I = 4 A

Substituting the values in the formula above we get,

V = 4 x 2 = 8 V

Hence, potential difference between the terminals of the battery = 8 V

(c) Current in 3 Ω resistor = ?

Using Ohm's law

V = IR

R = 3 Ω

V = 8 V

I = ?

Substituting the values in the formula above we get,

8 = I × 3

I = $\dfrac{8}{3}$

⇒ I = 2.66 A

Hence, current in 3 Ω resistor is 2.66 A

(d) Current in 6 Ω resistor = ?

Using Ohm's law

V = IR

R = 6 Ω

V = 8 V

I = ?

Substituting the values in the formula above we get,

8 = I × 6
⇒ I = 8 / 6 = 1.333 A

Hence, current in 6 Ω resistor is 1.34 A