Physics, asked by laskarranadeep, 1 day ago

FInd the power dissipated across R5

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Answered by prabhasaragada
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The power dissipated across resistor R5 in the network given below is

A. 3 watt B. 34 watt

C. 274 watt D. 9 watt

Explanation:

order to find power through a resistor, the required quantities are current through the resistor and the resistance of the resistor. By these, the power can be calculated by the Joule’s law of heating –

P=I2R

where I = current and R = resistance.

Step-by-step solution:

When 2 resistors are connected in series, the net resistance of the combination is the sum of the individual resistances.

R=R1+R2Question

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Related Questions

The power dissipated across resistor R5 in the network given below is

A. 3 watt B. 34 watt

C. 274 watt D. 9 watt

Answer

VerifiedVerified

46.8k+ views

6.4k+ likes

Hint:In order to find power through a resistor, the required quantities are current through the resistor and the resistance of the resistor. By these, the power can be calculated by the Joule’s law of heating –

P=I2R

where I = current and R = resistance.

Step-by-step solution:

When 2 resistors are connected in series, the net resistance of the combination is the sum of the individual resistances.

R=R1+R2

When 2 resistors are connected in parallel, the net resistance of the combination is the reciprocal of the sum of reciprocals of individual resistances.

1R=1R1+1R2

⇒R=R1R2R1+R2

Consider the arrangement of resistors connected in the given fashion, subject to the potential

difference of 12V.

In this circuit, the resistors are connectedfilledbookmark

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Grade 12

Sources Of Electricity

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Related Questions

The power dissipated across resistor R5 in the network given below is

A. 3 watt B. 34 watt

C. 274 watt D. 9 watt

Answer

VerifiedVerified

46.8k+ views

6.4k+ likes

Hint:In order to find power through a resistor, the required quantities are current through the resistor and the resistance of the resistor. By these, the power can be calculated by the Joule’s law of heating –

P=I2R

where I = current and R = resistance.

Step-by-step solution:

When 2 resistors are connected in series, the net resistance of the combination is the sum of the individual resistances.

R=R1+R2

When 2 resistors are connected in parallel, the net resistance of the combination is the reciprocal of the sum of reciprocals of individual resistances.

1R=1R1+1R2

⇒R=R1R2R1+R2

Consider the arrangement of resistors connected in the given fashion, subject to the potential difference of 12V.

In this circuit, the resistors are connected as follows:

The parallel combination of R3 is connected to R5. This combination is in series with R4. This entire combination is in parallel with R2. The above entire combination is in series with R1.

Hence, the net resistance R is given by –

R=R1+(R2//((R3//R5)+R4))

Let us substitute the values one-by-one and determine the resistance by applying the above formula for the net resistance.

R3//R5=R3R5R3+R5=3×63+6=189=2Ω

R3//R5+R4=2+2=4Ω

R2//((R3//R5)+R4)=4//4=4×44+4=168=2Ω

Finally, the net resistance is the series combination of R1 with the above combination.

R=R1+2=2+2=4Ω

The net current is given by –

I=VR=124=3A

In a parallel combination, the current branches depending on the resistances in the branches.

The current following through the combination

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