Question

compare the power used in length 2 ohm resistor in in each of the following circuits a 6 volt battery in series with one or more and 2 ohm resistor A 4 volt battery in parallel with 12 ohm and 2 ohm resistor​

Answer 1

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.

Answer 2

Answer:

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 -

R_{1} = 1 \: \& \: R_{2} = 2R

1

=1&R

2

=2

V [Potential Difference] = 6V

R_{1} + R_{2}⟹R 1 +R 2

1 + 2⟹1+2

3⟹3

From ohm's formula:

V = l \times R⟹V=l×R

I = \dfrac{V}{R}⟹I= RV

I = $\cancel\dfrac{6}{3}$ ⟹I= 3. 6

I = 2 A⟹I=2A

Now, finding power used:

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

Power = l^2 \times R⟹Power=l 2 ×R

P = 2^2 (2)⟹P=2

2

(2)

P = 4 (2)⟹P=4(2)

P = 8W⟹P=8W

Power is 8W in 2 Ω resistor.

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

V[Potential difference] = 4V

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

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}⟹P=

R

V

2

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

➡️P= 2 4 2

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

➡️P= 2

16

Power = 8 W➡️Power=8W

Power is 8W in 2 Ω resistor.

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