Question

Q.Three resistors of 4Ω, 6Ω and 8Ω are in parallel. In which resistor, power dissipation will be maximum? *
(1). 4Ω
(2). 6Ω
(3). 8Ω
(4). equal in all resistor

Answer 1

4) equal in all resisters

If the resisters are connected in parallel the power dissipated will be same in all resisters

If they are connected in series then,

P = V^2 / R

Maximum power will be dissipated in small resister i.e 4 Ω

Answer 2

Answer:

The power dissipation will be maximum in 4Ω resistor.

Explanation:

Given the resistors of resistances,

$R_1=4\Omega, R_2=6\Omega~ \mbox{and}~ R_3=8\Omega$

The power dissipated by each resistor connected in parallel in the circuit is given by

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

where V is the voltage across the resistor and R is the resistance of the resistor.

Since the resistors are connected in parallel, therefore the same voltage is shared by all the resistors.

Let the common voltage across the resistors be V volts.

Thus, the ratio of the power dissipation by the resistors can be given by

$P_1:P_2:P_3=\dfrac{V^{2} }{R_1}: \dfrac{V^{2} }{R_2}:\dfrac{V^{2} }{R_3}$

                  $=\dfrac{1 }{4}: \dfrac{1}{6}:\dfrac{1}{8}=6:4:3$

Therefore, $P_1:P_2:P_3=6: 4:3$

So, the power dissipation will be maximum in the first resistor, 4Ω

So, the correct answer is option 1.

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