Question

Electric device works properly if the P.D. across it is 110 V, when the current through it is found
to be 2.4 A. What resistance must be connected in series with it if is to run on a 230 V source?
plz answer fast
points for the best answer​

Answer 1

Answer:

Given:

When voltage = 110 V, then safety current = 2.4 A

Now , 230 V of Potential difference is applied across the circuit.

To find:

The resistance to be connected in series for the device to work safely

Concept:

The resistance in series keeps the current in a safety limit. This helps in the proper functioning of the device.

Calculation:

Resistance of the device = V/I

=> R1 = 110/(2.4)

=> R1 = 45.833 ohm.

Now, let the resistance to be added in series be R2

So,

V/(R1 + R2) = (safety current)

=> 230/(45.833 + R2) = 2.4

=> 45.833 + R2 = 230/2.4 = 95.833

=> R2 = 50 ohm

So, 50 ohm resistor has to be added in series to keep the device in safe conditions

Answer 2

Question :----- we have to find what resistance must be connected in series to work it safely ....

Given:----------

• potential difference = 110V
• current = 2.4 A
• After that 230V of PD applied .

Solution :-----

we know that resistance is the opposition to current flow.

According to Ohm's Law the relationship between the voltage (V), current (I) and resistance (R) is :--------

V = I × R

So,

R = V/I

R = 110/2.4 = 1100/24 = 275/6 = 45.833 Ω

Now, By voltage division Formula we know that,

V/(R1+R2) = Safety current ,

Given New V = 230V , Safety current = 2.4 , R1 = 45.833 Ω

Putting all values now , we get,

230/(45.833+R2) = 2.4

→ 230/2.4 = 45.833 + R2

→ 2300/24 = 45.833 + R2

→ 575/6 = 45.833 + R2

→ 95.833 = 45.833 + R2

→ R2 = 95.833 - 45.833

→ R2 = 50 Ω

So, 50 Ω Resistance is Required in series to work it safely.