Question

Find the power dissipated when:

(a) a current of 5 mA flows through a resistance of 20 k

(b) a voltage of 400 V is applied across a 120 k resistor

(c) a voltage applied to a resistor is 10 kV and the current flow is 4 mA.​

Answer 1

Answer:

Explanation:

it is given in the question that a current of $5 mA$ flows through a resistance of $20 k$ and we need to find the power dissipated and (b) a voltage of $400 V$ is applied across a $120 k$ resistor and the last question is  a voltage applied to a resistor is $10 kV$ and the current flow is $4 mA$.​

1) in the first question current is $I =5mA$ converting it into ampere we get the answer

$=0.005A$

the resistance is

$r= 20$ ohm

converting it to ohms we get

=$20000$ohm

putting the voltage formula

where voltage is equal to current multiplies by resistance

V=IR

putting the values given in the question

v=$0.005 X20000$

hence the answer is

$v=$$100v$

we have got the value of voltage , let us put the value in power formula

where power is equal to voltage multiplied by current

p=vi

putting the value we get

$p=100vX0.005A$

hence the answer is power equals to

$p=0.5W$

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Answer 2

Complete Question: Find the power dissipated through the resistor when:

(a) A current of 5 mA flows through a resistance of 20 k Ω

(b) A voltage of 400 V is applied across a 120 k Ω resistor

(c) A voltage applied to a resistor is 10 kV and the current flow is 4 mA

Solution:

→ When a current flows through a resistor, then some of the electrical Power gets dissipated in the form of heat energy. Every resistor with a potential drop across it will lose some of the electrical power in the form of Heat energy to the surrounding.

→ The power rating of a resistor is equal to the maximum power that can be dissipated from a resistor without getting burned out.

→ The power dissipated (P) by a resistor (R) if the potential difference across the resistor is 'V' and the current through the resistor is 'I' is given as:

                     $Power\;dissipated(P)=I^{2}R=VI=\frac{V^{2} }{R}$

(a) Current (I) = 5 mA = 5 × 10⁻³ A

    Resistance (R) = 20 k Ω = 20 × 10³ Ω

 → Power dissipated (P) = I²R

                                ∴ P = (5 × 10⁻³)² ×(20 × 10³)

                                ∴ P = 25 × 10⁻⁶ × 20 × 10³

                                ∴ P = 0.5 Watt

→ Therefore the power dissipated is equal to 0.5 Watt.

(b) Voltage (V) = 400 V

    Resistance (R) = 120 k Ω = 120 × 10³ Ω

     $Power\: dissipated(P)=\frac{V^{2} }{R}=\frac{400^{2} }{120\times10^{3} } \\\\\therefore P = 1.33\:Watt$

→ Therefore the power dissipated is equal to 1.33 Watt.

(c) Voltage (V) = 10 kV = 10⁴ V

    Current (I) = 4 mA = 4 × 10⁻³ A

→  Power dissipated (P) = V × I

                                ∴ P = (10⁴)×(4 × 10⁻³)

                                ∴ P = 40 Watt

→ Therefore the power dissipated is equal to 40 Watt.

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