FInd the power dissipated across R5
Answers
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
Answers
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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CBSE
Physics
Grade 12
Sources Of Electricity
Question
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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