Question

(a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistanceof a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.(b) In an electric circuit two resistors of 12 Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery.

Answer 1

B)1/R=1/12+1/12
R=6ohms
V=IR
I=6/6=1A
Hope it helps

Answer 2

(a) Equivalent resistance for parallel resistors:

The parallel resistors connection is given in the image below.

Let the resistors R1, R2, R3 ... Rn are connected in parallel connection with emf E.

Since, the connection is parallel the potential difference across the circuit remains the same.

Let the potential difference across R1, R2, R3 ... Rn be V.

Let the current flowing through R1, R2, R3 ... Rn be I1, I2, I3 ... In.

Let $\bold{R_{eq}}$ be the equivalent resistance of the resistors.

According to Ohm's law, we get,

$\bold{E=I \times R_{e q} \longrightarrow(1)}$

Current I1 for resistor R1 is:

$\bold{I_{1}=\frac{E}{R_{1}}}$

Current I2 for resistor R2 is:

$\bold{I_{2}=\frac{E}{R_{2}}}$

Now, current for resistor Rn is:

$\bold{I_{n}=\frac{E}{R_{n}}}$

The total current flowing through the circuit is total of all the current flowing through the circuit.

$\bold{I=I_{1}+I_{2}+\ldots \ I_{n}}$

On substituting the current in above equation, we get,

$\bold{\frac{E}{R_{e q}}=\frac{E}{R_{1}}+\frac{E}{R_{2}}+\ldots \ \frac{E}{R_{n}}}$

$\bold{\therefore \frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots \ \frac{1}{R_{n}}}$

Thus, the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances. Hence proved.

(b) The current drawn from the battery is 6 Ω.

Given:

Resistance of two resistors = 12 Ω = R1 = R2

Potential difference = 6 V = V

To find:

Current drawn = I = ?

Formula:

$\bold{\frac{1}{R_{net}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}}$

Solution:

Since, the resistors are connected in series. The net resistance is equal to reciprocal of the individual resistance.

On substituting the value of the resistance, we get,

$\bold{\frac{1}{R_{n e t}}=\frac{1}{12}+\frac{1}{12}}$

$\bold{\frac{1}{R_{n e t}}=\frac{2}{12}=\frac{1}{6}}$

$\bold{R_{net}=6 \ \Omega}$