Question

compare the power used in 2 ohm resistors in each of the following in of the following circuits
1. a 6 volt battery in series with 1 ohm 2 ohm resistor
2. A 4 volt battery in parallel with 12 ohm and 2 ohm resistor

Answer 1

r= 2 ohm
p.d = 6v
$p = {v}^{2} \div r \\ p = {6 }^{2} \div 2 \\ p = 36 \div 2 \\ p = 18w$
now for the second one

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.