Question

Please solve 2, 3 and 4 with workouts.............​

Answer 1

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Resistance R1 of the bulb is given by the expression,

Supply voltage, V = 220 V

Maximum allowable current, I = 5 A

Rating of an electric bulb P=10watts

Because R=V2/P

According to Ohm’s law,

V = I R

Let R is the total resistance of the circuit

for x number of electric bulbs

R=V/I

=220/5=44 Ω

Resistance of each electric bulb,

R1=4840Ω

∴ Number of electric bulbs connected in parallel are 110.

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⠀⠀⠀2. Supply voltage, V = 220 V

Here Resistance of one coil, R =24 Ω

(i) Coils are used separately

According to Ohm’s law, V=I1R1

I1 is the current flowing through the coil

I1=V/R1=220/24=9.166A

∴ current flow through the coil when used separately is 9.16 A.

(ii) Coils are connected in series

R2=24 Ω+24 Ω=48 Ω

According to Ohm’s law

I2=V/R2=220/48=4.58A

the current flowing through the series circuit is 4.58A

(iii) Coils are connected in parallel

According to Ohm’s law

I3=V/R3=220/12=18.33A

the current flowing through the parallel circuit is 18.33A

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⠀⠀3. (I) Given :-

Potential difference = 6 V

Solution :-

Equivalent resistance = 1 + 2 = 3 Ω

According to the Ohm’s law

V = IR

I = V/R

I = 6/3

I = 2 A

Current through the circuit = 2 A

we know that Power = I²R

P = 2 × 2 × 2

P = 8 W

Hence, the power used by 2 Ω resistor is 8 W.

(ii)

Given :-

Potential difference, V = 4 V

Voltage across each component of a parallel circuit remains the same. Then,

Solution :-

P = V²/R

P = 4²/2

P = 8 W

Hence, the power used by 2 Ω resistor is 8 W.