Question

A current of 5 A flows in the winding of an electric motor, the resistance of
the winding being 100 Ω. Determine a) the p.d. across the winding and
b) the power dissipated in the coil.​

Answer 1

Answer:

Explanation:

ANSWER:

Given the following values:-

CURRENT = 5 Amperes

RESISTANCE = 100Ω

POTENTIAL DIFFERENCE / VOLTAGE = ?

POWER = ?

For attempting this Question, we need to remember two formulas

• $V = IR$
• $P = VI$

Where :-

1. V is Voltage or Potential Difference
2. I is current
3. R is resistance offered
4. P is power.

Finding the Voltage first,

a) We know the formula V = IR

Placing I and R in the values we get

V = 5*100

V = 500 volts

Now we got the potential difference as 500 volts.

b) We need to find power:-

We know the formula P = VI

Placing the values of V and I , we get

P = 500*5

P = 2500 Watts.

So power is 2500 watts.

Things to Note:-

• Remember the Formulas given above
• Maintain the SI Units of the following
• (p.d)Potential difference /Voltage - Volts(V)
• Current -  Amperes(A)
• Resistance - Ohm(Ω)
• Power - Watts(W)

If any value is not given in SI unit, convert it and then calculate.

Answer 2

Given :

• Current Flowing in motor (I) = 5 A
• Resistance (R) = 100 Ω

To Find :

• Potential Difference across the winding
• Power Dissipated in the coil

Formulae Used :

• V = IR
• P = VI

Explanation :

We are given current flowing in the motor is 5 Ampere (A) , and Resistance is given as 100 ohm (Ω). We have to find out Potential difference across the winding and power Dissipated in the coil.

In this question we have to use two formulae, First one V = IR. This formula is also known as Ohm's Law :

Where,

• V is potential difference
• I is current
• and R is resistance

Another formula is P = VI, by using this we will calculate the power dissipated in the coil.

Where,

• P is power
• V is potential Difference
• I is current

Solution :

Use Ohm's Law :

$\implies \sf{V \: = \: IR} \\ \\ \implies \sf{V \: = \: 5 \: \times \: 100} \\ \\ \implies \sf{V \: = \: 500}$

$\therefore$ Potential difference across the winding is 500 V.

___________________________

Now, we will calculate Power

$\implies \sf{P \: = \: VI} \\ \\ \implies \sf{P \: = \: 500 \: \times \: 5} \\ \\ \implies \sf{P \: = \: 2500}$

$\therefore$ Power Dissipated in the coil is 2500 Watts.