Question

In the circuit diagram given below, find (i) total resistance of the circuit, (ii) total current flowing in the circuit, (iii) the potential difference across R1 and (iv) the potential difference across R2.​

Answer 1

Given :

In the circuit,

• R₁ = 8 ohms.
• R₂ = 4 ohms.
• R₃ = 4 ohms.

Potential difference : 6V.

To find :

(i) total resistance of the circuit,

(ii) total current flowing in the circuit,

(iii) the potential difference across R₁ and

(iv) the potential difference across R₂.

Solution :

Here R₂ and R₃ are connected in parallel combination

We know that,

The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

1/R = 1/R₁ + 1/R₂

By substituting the values in the formula,

$\dashrightarrow \sf \dfrac{1}{R_{23}} = \dfrac{1}{R_{1}} + \dfrac{1}{ R_2}$

$\dashrightarrow \sf \dfrac{1}{R_{23}} = \dfrac{1}{4} + \dfrac{1}{4}$

$\dashrightarrow \sf \dfrac{1}{R_{23}} = \dfrac{1 + 1}{4}$

$\dashrightarrow \sf \dfrac{1}{R_{23}} = \dfrac{2}{4}$

$\dashrightarrow \sf \dfrac{1}{R_{23}} = \dfrac{1}{2}$

$\dashrightarrow \sf R_{23} = 2 \: ohms$

Now, R₁ and R₂₃ are connected in series combination,

We know that,

» The combined resistance of any number of resistance connected in series is equal to the sum of the individual resistances. i.e.,

R = R₁ + R₂

By substituting the values in the formula,

$\dashrightarrow \sf R_{eq} = R_{1} + R_{23}$

$\dashrightarrow \sf R_{eq} = 8 + 2$

$\dashrightarrow \sf R_{eq} = 10 \: ohms$

Thus, the total resistance of the circuit is 10 ohms.

Now, using Ohms law that is,

» At constant temperature the current flowing through a conductor is directly proportional to the potential difference across its ends.

Formula : V = RI

where,

• V denotes potential difference
• R denotes resistance
• I denotes current

substituting all the given values in the formula,

$\dashrightarrow \sf V = RI$

$\dashrightarrow \sf 6 = (10)I$

$\dashrightarrow \sf I = \dfrac{6}{10}$

$\dashrightarrow \sf I = 0.6A$

Thus, total current flowing in the circuit is 0.6 A.

The potential difference across R₁ is,

$\dashrightarrow \sf V_1 = R_1I$

$\dashrightarrow \sf V_1 = 8 \times 0.6$

$\dashrightarrow \sf V_1 = 4.8 \: volts$

Thus, the potential difference across R₁ is 4.8 volts.

The potential difference across R₂ is,

$\dashrightarrow \sf V_2 = R_{23}I$

$\dashrightarrow \sf V_2 = 2 \times 0.6$

$\dashrightarrow \sf V_2 = 1.2 \: volt$

Thus, the potential difference across R₂ is 1.2 volts.