Question

Three resistors are connected as shown in the diagram.

Through the resistor 5 ohm , a current of 1 ampere is flowing.
(i) What is the current through the other two resistors?
(ii) What is the p.d. across AB and across AC?
(iii) What is the total resistance?

Answer 1

Given,

Current in the circuit is 1 A

Two resistors across BC are in parallel combination

So, Equivalent resistance across BC

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

Or,

$\frac{1}{R_e}=\frac{1}{15}+\frac{1}{10}$

$\frac{1}{R_e}=\frac{1}{6}$

Therefore,

$R_e=6$

Now this resistor is in series combination with the 5 ohm resistor

So, total resistor of the circuit is

5 ohm + 6 ohm = 11 ohm

Now,

Applying Ohm's law

$V=IR$

Or,

$V=1\times 11$

$V=11V$

Hence, total voltage in the circuit is 11 V

Potential difference across the 5 ohm resistance is 5 V

So, Potential difference across the BC will be 6 V

(i) Current through the 10 ohm resistor (I) = $\frac{V}{R}=\frac{6}{10}=\frac{3}{5}A$

Current through the 15 ohm resistor = $\frac{V}{R}=\frac{6}{15}=\frac{2}{5}A$

(ii) Potential difference across AB

$V=IR=1\times 5=5V$

Potential difference across BC = Total voltage - Voltage across AB

Or, 11 V - 5 V = 6 V

(iii) Total resistance

As calculated above total resistance of the circuit is 11 ohm.