Question

compare the power used in the 2 ohm resistor in each of the following circuit : 6 volt battery in series with a 1 ohm and 2 ohm resistors and a 4 volt battery in parallel with 12 ohm and 2 ohm resistors​

Answer 1

Explanation:

(i) Potential difference, V = 6 V

1Ohm and 2 Ohm resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ohm

According to Ohm’s law,

V = IR

Where,

I is the current through the circuit

I= 6/3 = 2 A

This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2Ohm resistor is 2 A. Power is given by the expression,

P= (I)2R = (2)2 x 2 = 8 W

(ii) Potential difference, V = 4 V

12 Ohm and 2 Ohm resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ohm resistor will be 4 V.

Power consumed by 2 Ohm resistor is given by

P= V2/R = 42/2 = 8 W

Therefore, the power used by 2 Ohm resistor is 8 W.

Answer 2

Answer:

Correct Question:

Compare the power used in length 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

AnswEr :

i) A 6 V Battery in series with 1 Ω & 2 Ω resistors -

$\sf R_{1} = 1 \: \& \: R_{2} = 2$

V [Potential Difference] = 6V

$\implies\sf R_{1} + R_{2}$

$\implies\sf 1 + 2$

$\implies\sf 3$

From ohm's formula:

$\implies\sf V = l \times R$

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

$\implies\sf I = \cancel\dfrac{6}{3}$

$\implies\sf I = 2 A$

Now, finding power used:

[We've to find power in 2 Ω resistor.]

$\implies\sf Power = l^2 \times R$

$\implies\sf P = 2^2 (2)$

$\implies\sf P = 4 (2)$

$\implies\sf P = 8W$

Power is 8W in 2 Ω resistor.

$\rule{150}2$

ii) 4 V battery in parallel with 12 Ω and 2 Ω resistors.

V[Potential difference] = 4V

$\sf R_{1} = 12 \: \& \: \: R_{2} = 2$

[When both the resistors connected in ||s, their voltage of parallel Circuits remains same.]

Finding power in 2 Ω resistor:

$\implies\sf P = \dfrac{V^2}{R}$

$\implies\sf P = \dfrac{4^2}{2}$

$\implies\sf P = \cancel\dfrac{16}{2}$

$\implies\sf Power = 8 W$

Power is 8W in 2 Ω resistor.

$\rule{150}2$

In ohm's formula, V is Voltage, I is current & R is resistor.