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Question: If power consumptions in \(R_1\)​, \(R_2\)​ & \(R_3\)​ are the same then what will be the relation b...

If power consumptions in R1R_1​, R2R_2​ & R3R_3​ are the same then what will be the relation between them?

A. R1=R2=R3{R_1} = {R_2} = {R_3}
B. R2=R3=2R1{R_2} = {R_3} = 2{R_1}
C. R2=R3=4R1{R_2} = {R_3} = 4{R_1}
D. R1=R2=4R3{R_1} = {R_2} = 4{R_3}

Explanation

Solution

As we all know that the power consumption in an electrical circuit is the product of the voltage and the current flowing in the circuit but power consumption in a circuit in the terms of the resistance will be P=V2RP = \dfrac{{{V^2}}}{R}, and also P=I2RP = {I^2}R. Hence from this relation we can conclude that the power consumption in a circuit is dependent on the voltage and the resistance and also on the current.

Complete answer:
Now on observing the given circuit we see that the resistors R2{R_2} and R3{R_3} are connected in parallel to each other and the combination of both of them is connected in series with the resistance R1{R_1}. Now if we assume that the current II is flowing through the resistance R1{R_1} and current I1{I_1} is flowing through resistance R2{R_2}, then the current flowing through the resistance R3{R_3} will be II1I - {I_1}.
Now applying current divider rule,
I1=I×R3R2+R3{I_1} = I \times \dfrac{{{R_3}}}{{{R_2} + {R_3}}}-----equation (1)
It is given in the question that the power across all the resistors are the same, that is P1=P2=P3{P_1} = {P_2} = {P_3}.
Now if P1=P2{P_1} = {P_2},
Then I2R1=I12R2{I^2}{R_1} = {I_1}^2{R_2}-------equation (2)
Also P2=P3{P_2} = {P_3}
Voltage across R2{R_2} and R3{R_3}will be the same.
V2R2=V2R3\Rightarrow \dfrac{{{V^2}}}{{{R_2}}} = \dfrac{{{V^2}}}{{{R_3}}}
R2=R3\Rightarrow {R_2} = {R_3}--------equation (3)
Now putting the values from equation (3) in equation (1), we get
I1=I×R32R3{I_1} = I \times \dfrac{{{R_3}}}{{2{R_3}}}
I1=I2\Rightarrow {I_1} = \dfrac{I}{2}
Now putting the values in equation (2), we get
I2R=I12R2{I^2}R = {I_1}^2{R_2}
I2R=[I2]2R2\Rightarrow {I^2}R = {\left[ {\dfrac{I}{2}} \right]^2}{R_2}
I2R1=I24R2\Rightarrow {I^2}{R_1} = \dfrac{{{I^2}}}{4}{R_2}
R1=R24\Rightarrow {R_1} = \dfrac{{{R_2}}}{4}
4R1=R2\Rightarrow 4{R_1} = {R_2}----equation (4)
Now combining equation (3) and equation (4), we get
R2=R3andR2=4R1\Rightarrow {R_2} = {R_3}and{R_2} = 4{R_1}
R2=R3=4R1\Rightarrow {R_2} = {R_3} = 4{R_1}

So, the correct answer is “Option C”.

Note:
The power consumption in the series combination is less than the parallel combination. For example, if we have to connect 12 LEDs of 25mA each across a 3V supply and we connect then in parallel then the consumed power will be 0.9 watts while if we connect the LEDs in the series connection then the power consumed will be 0.7watt. but the bulbs will glow brighter in the parallel connection while the bulbs in the series connections are dim. So finally, the parallel connection is better even though it consumes comparatively more power.