Question

Please solve this

Class 10th

Answer 1

4): For x number of resistors of resistance 176 Ω,

Here Supply voltage, V = 220 V

Current, I = 5 A

Equivalent resistance of the combination = R, given as

From Ohm’s law,

∴four resistors of 176 Ω are required to draw the given amount of current.

6):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.

7):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

$\eqalign{ & {1 \over {{R_3}}} = {1 \over {24}} + {1 \over {24}} \cr & {1 \over {{R_3}}} = {{1 + 1} \over {24}} \cr & {1 \over {{R_3}}} = {2 \over {24}} = {1 \over {12}} \cr & \Rightarrow {R_3} = 12\Omega \cr} $

According to Ohm’s law

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

the current flowing through the parallel circuit is 18.33A