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Answer 1

» Required Answers:

20(i) The value of current through 5Ω resistor:

$\large \boxed { \bf \red { Option \: B ( 1.2A)}}$

Here we have to find the value of current through 5Ω resistor,

We know that,

• $\tt { \dfrac{V}{I}=R}$

Where,

• V ( potential difference) = 6V
• I (current) =?
• R (resistance) = 5Ω

Therefore:

$\tt { \dfrac{V}{I}=R}$

$\tt { \leadsto \dfrac{6}{I}=5}$

$\tt { \leadsto 5I=6}$

$\tt { \leadsto I= \dfrac{6}{5} }$

$\tt \red { \leadsto I= 1.2A}$

20(ii) The total current in the circuit :

$\large \boxed { \bf \green {Option \: B!! ( 2A)} }$

To find the value of current (I) we should know the total resistance (R) of the circuit. Here we have 3 resistors 5Ω, 10Ω and 30Ω in parallel.

Total resistance = $\tt { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}+ \dfrac{1}{R_3}}$

$\tt { \implies \dfrac{1}{R_p} = \dfrac{1}{5} + \dfrac{1}{10}+ \dfrac{1}{30}}$

$\tt { \implies \dfrac{1}{R_p} = \dfrac{6+3+1}{30} }$

$\tt { \implies \dfrac{1}{R_p} = \dfrac{10}{30} }$

$\tt { \implies \dfrac{1}{R_p} = \dfrac{1}{3} }$

$\tt \purple { \implies R_p= 3 Ω }$

Now,

• Total resistance (R) =3 Ω
• Potential difference (V) = 6V
• Current (I) =?

Applying Ohm's law to the whole circuit, we get:

• $\tt { \dfrac{V}{I}=R}$

So that,

$\tt { \dfrac{V}{I}=R}$

$\tt { \leadsto \dfrac{6}{I}=3}$

$\tt { \leadsto 3I=6}$

$\tt { \leadsto I= \dfrac{6}{3} }$

$\tt\red { \leadsto I= 2 A }$

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20 (iii) Total effective resistance of the circuit:

$\large \boxed { \bf \green {Option \: D!! ( 3Ω)} }$

• R1 = 5Ω
• R2 = 10Ω
• R3 = 30Ω

Since the resistors are in parallel combination, therefore,

Total effective resistance = $\tt { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}+ \dfrac{1}{R_3}}$

$\tt { \implies \dfrac{1}{R_p} = \dfrac{1}{5} + \dfrac{1}{10}+ \dfrac{1}{30}}$

$\tt { \implies \dfrac{1}{R_p} = \dfrac{6+3+1}{30} }$

$\tt { \implies \dfrac{1}{R_p} = \dfrac{10}{30} }$

$\tt { \implies \dfrac{1}{R_p} = \dfrac{1}{3} }$

$\tt \purple { \implies R_p= 3 Ω }$

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20 (iv) The value of current through 10 Ω resistor :

$\large \boxed { \bf \green {Option \: C!! ( 0.6A)} }$

Here we have to find the value of current through 10 Ω resistor,

We know that,

• $\tt { \dfrac{V}{I}=R}$

Therefore,

$\tt { \dfrac{V}{I}=R}$

$\tt { \leadsto \dfrac{6}{I}=10}$

$\tt { \leadsto 10I=6}$

$\tt { \leadsto I= \dfrac{6}{10} }$

$\tt { \leadsto I= \dfrac{3}{5} }$

$\tt \red { \leadsto I= 0.6A}$

Therefore,the value of current through 10 Ω resistor is 0.6 Ampere.

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20 (v) The value of total current in the circuit we replace 6V battery by 9V battery:

$\large \boxed { \bf \green {Option \: A!! ( 3A)} }$

Here, potential difference will only change.We have found the resistance in the second part of the question. Now we have

Now,

• Total resistance (R) =3 Ω
• Potential difference (V) = 9V
• Current (I) =?

Applying Ohm's law to the whole circuit, we get:

$\tt { \dfrac{V}{I}=R}$

So that,

$\tt { \dfrac{V}{I}=R}$

$\tt { \leadsto \dfrac{9}{I}=3}$

$\tt { \leadsto 3I=9}$

$\tt { \leadsto I= \dfrac{9}{3} }$

$\tt\red { \leadsto I= 3A }$

Therefore, The value of total current in the circuit we replace 6V battery by 9V battery will be 3A.

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