Question

Two resistors 10 and 20 are connected in series across a cell of emf 3V. Find (i) the current flowing in the circuit and (ii) the potential difference across 10 Ω resistor.

Answer 1

★ Given :-

• R 1 = 10 Ω
• R 2 = 20 Ω
• EMF = 3 V

★ To find :-

•  Current flowing through the circuit
• Potential difference across 10 Ω resistor

★ Solution :-

First we need to find the equivalent resistance of the circuit

Equivalent resistance for two resistors connected in Series is given by the formula ,

$\bigstar\boxed{\mathtt{ R_s = R_1 + R_2 ....+ R_n }}$

hence ,

$\implies\mathtt{ R_s = R_1 + R_2 }$

$\implies\mathtt{ R_s = 10 + 20}$

$\implies\mathtt{ R_s = 30 \Omega}$

The equivalent resistance of the circuit is 30Ω

Now let's find the net current flowing through the circuit in order to do that simply apply ohm's law

According to ohm's law ,

$\bigstar\boxed{\mathtt{ V= IR }}$

hence ,

$\implies\mathtt{ I = \dfrac{V}{R} }$

$\implies\mathtt{ I = \dfrac{3}{30} }$

$\implies\mathtt{ I =0.1 A }$

The current flowing through the circuit is 0.1 A

We know that when two resistors are connected in series with each other the potential difference across both the resistors is different but the current flowing through both these resistors is same

Potential difference across 10Ω resistor

$\implies\mathtt{ V = I R }$

$\implies\mathtt{ V = 0.1 \times 10 }$

$\implies\mathtt{ V = 1 V }$

The potential difference across 10Ω resistor is 1 volt