Question

[Physics]

Two resistors A and B of resistance 4 ohm and 6 ohm respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate :
(i) the power supplied by the battery
(ii) the power dissipated in each resistor.
​

Answer 1

Answer:-

• Power supplied by battery= 15 watt

• Power decipated by resistor A = 9watt

• Power decipated by resistor B = 6 watt

Given:

• Resistor A = 4Ω

• Resistor B = 6Ω

• Voltage across terminal of battery = 6volt

To find:

• Power supplied by battery

• Power decipated in each resistor

Step-by-step explanation:

Firstly we have to find total resistance of circuit so that we could find the power supplied by battery.

Resistance in parallel connection can be expressed as::

→ 1/R = 1/R₁ + 1/R₂ + 1/R₃...

→ 1/R = 1/4Ω + 1/6Ω

→ 1/R = (3+2)/12Ω

→ 1/R = 5/12Ω

→ R = 12/5Ω

_______________________

~Finding power supplied by battery :-

We have formula for power::

→ P = V²/R

Here,

• P = Power

• V = Voltage by battery

• R = Equivalent resistance of circuit

Now solving:-

→ P = (6)²/(12/5)

→ P = (36)/(12/5)

→ P = (36×5)/12

→ P = (3×5)

→ P = 15 watt

_______________________

~Finding power decipated by resistor A

Applying formula for power::

→ P₁ = V²/Rₐ

→ P₁ = (6)²/4Ω

→ P₁ = 36/4

→ P₁ = 9 watt

_______________________

~Finding power decipated by resistor B

Applying formula for power::

→ P₂ = V²/R$\sf_b$

→ P₂ = (6)²/6

→ P₂ = 36/6

→ P₂ = 6 watt

Answer 2

Given :-

• Two resistors are A and B are connected in parallel

• Resistance of Resistor A = 4 ohm

• Resistance of Resistor B = 6 ohm

• The combination is connected across a 6V battery.

Solution :-

1) Two resistors are connected in parallel

Therefore,

As we know that,

Equivalent resistance in parallel = 1/R in parallel

Subsitutes the required values,

1/R in parallel = 1/R of A + 1/R of B

1/R in parallel = 1/4 + 1/6

1/R in parallel = 6 + 4 / 24

1/R in parallel = 10/24 ohm

R = 2.4 ohm

Thus, Equivalent resistance is 2.4 ohm

Now,

Power = VI

P = IR * I [ V = IR ]

P = I^2R. [ I = V/R ]

P = V/R * V/R * R

P = V^2/R

Subsitute the required values,

P = 6 * 6 / 2.4

P = 36/2.4

P = 15W

Hence, The power supplied by a battery is 15W

2) Power dissipated across each resistor is given by , P = VI

Now,

Currect across Resistor A

I = V/R

I = 6/4

I = 1.5A

Power dissipated in resistor A

P = VI

P = 6 * 1.5

P = 9W

Now,

Current across Resistor B

I = V/R

I = 6/6

I = 1 A

Power dissipated in Resistor B

P = VI

P = 6 * 1

P = 6W

Hence, The power dissipated in each resistor is 9W and 6W $.$