Question

For the circuit shown in the diagram below
What is the value of
(i) current through 6 Ω resistor?
(ii) potential difference across 12 Ω resistor?

Answer 1

Follow Ohm's law

Explanation:

• The resistance of $6 \Omega$ and $3 \Omega$ are connected in series.  

Therefore, their net resistance can be calculated as

$R = R_1 + R_2$

Here, $R_1 = 6 \Omega$

$R_2 = 3 \Omega$

So, R = 6 + 3 = $9 \Omega$

• The current through 6 ohm resistor= current through line 1,  

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

= $\frac {4}{9}$ = 0.44 A

(ii) The resistance of $12 \Omega$ and $3 \Omega$ are connected in series. Therefore, their net resistance can be calculated as

$R = R_1 + R_2$

Here, $R_1 = 12 \Omega$

& $R_2 = 3 \Omega$

So, R = 12 + 3 = $15 \Omega$

Current through them = $I = \frac {V}{R} = \frac {4}{15}$

= $\frac {4}{15} \times 12$

= 3.2 A