# Heating effect of electric current class 10 Numericals

Last updated on August 25th, 2023 at 02:19 pm

In this post on the *Heating effect of electric current class 10 Numericals*, we will list down the *Heating Effect formulas or Equations* of electric power and energy. And, then we will solve a bunch of numerical problems as well using these formulas.

**Formulas Used**

- The power P dissipated in a component due to the heating effect of current is related to the potential difference (PD) V across the component and the current I in it:
**P = IV** - The energy E converted in time Δt is
**E = IVΔt**

When either V or I is unknown, then a few more equations become available. And these are as follows:

- Formulas of Power
P = IV

When V is not known:P =IV=I IR= I^{2}R

When I is not known:P = IV =[V/R]V= V^{2}/R- Formulas of Energy
E = IV Δt

When V is not known:E = IV Δt = I[IR]Δt= I[ Using Ohm’s Law equation}^{2}R Δt

When I is not known:E = IV Δt =[V/R]VΔt= (V^{2}/R) Δt

The equation of energy E given above will be used to find out the heat energy or heat generated due to the current flow in a conductor.This means:H =I^{2}R Δt

[ Note: the related law is known asJoule’s law of heating.

These equations will allow you to calculate the energy converted into electrical heaters and lamps and so on. Applications that you may come across include heating calculations, and determining the consumption of energy in domestic and industrial situations.

**Numericals on the heating effect of current – solved [questions and answers]**

1) Calculate the power dissipated in a 250 Ω resistor when the PD across it is 10 V.

**Solution**

P = V^{2}/ R

= 10^{2}/ 250

= 0.40 W

2 ) A 9.0 kW electrical heater for a shower is designed for use on a 250 V mains supply. Calculate the current in the heater.

**Solution**

P = IV

=> I = P/V

=> I= 9000/250

=> I= 36 A

3 ) Calculate the resistance of the heating element in a 2.0 kW electric heater that is designed for a 110 V mains supply.

**Solution**

P = V^{2}/R=> R = V

^{2}/P=> R= 110

^{2}/ 2000=> R= 6.1 A.

4] How much will be the heat generated if 0.2 A current flows through a conductor of resistance 200 ohms for 5 minutes?

**Solution:**

I = 0.2 A, R=200 ohm, **Δt** = 5 min = 300 s

According to Joule’s Law, **H =** ** I ^{2}R Δt **= [0.2]

^{2}. 200. 300 = 2400 J

So, amount of heat generated: 2400 J

*[ In cal it would be 2400/4.2 cal = 571.43 cal. Note: 4.2 J = 1 calorie.]*

5] Find out the heat developed in 3 minutes by a device of resistance 920 ohms working under 230 V.

**Solution:**

We can use this formula **H =** **(V ^{2}/R) Δt**

V = 230 V, R = 920 ohms, **Δt**= 3 min = 180 s

H =(V= [230^{2}/R) Δt^{2}/920]. 180 =10350 J

**Practice Problems**

- A bulb is connected across a 12 V battery; the resistance of the bulb filament is 4 Ω. Calculate the energy transferred to the bulb in 10 minutes.
- Compute the heat generated when a current of 10 A passes through a heater coil for 1hr when connected across the line voltage of 220 V.
- A battery sends a current of 3 A through a 5 Ω resistor. Find the electric energy supplied by the battery in 15 minutes.

**Solution of practice problems:**

1 ) Battery voltage, V = 12 V.

Resistance of the bulb, R = 4 Ω.

Time, t = 10 min = 600 s.

Current flowing through filament, I is obtained by applying Ohm’s Law.

I = V/R = 12 V/4 Ω = 3 A.

Energy transferred to the bulb in time t is =I^{2}R t.

Using the above values, energy transferred= (3)^{2}× (4) × (600) = 21,600 J.

2 ) Current I = 10 A.

Potential difference, V = 220 V.

Time, t = 1hr = 3600 s.

We know that the heat produced in time t when current I flows through a potential difference V is

H = V I t

So, H = (220 V) × (10 A) × (3600 s)

= 7.92 × 10^{6}J.