Question

If the resistors of resistances 10 Ω, 20 Ω and 60 Ω are connected in parallel to a battery of 12 V and of negligible internal resistance, the effective resistance of the circuit will be …....​

Answer 1

Answer:

Solution

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The current in the 10Ω resistance is

i

1

=

R

1

+R

2

R

2

i=

20+10

(20)×(6)

=4A

The current in the 20Ω resistor is

i

2

=

R

1

+R

2

R

1

i=

10+20

10×6

=2A

Or

emf across 10Ω resistor

V

1

=i

1

×10

emf across 20Ω resistor

V

2

=i

2

×20

Since i

1

+i

2

=6A and V

1

=V

2

, we have i

1

=4A

i

2

=2A

Answer 2

Explanation:

Given:

Three resistance in parallel $10$Ω, $20$Ω and $60$Ω.

Potential difference $12V$.

Internal resistance of battery is considered to be 0.

To find: Effective resistance

Calculation:

All the individual resistances will have same potential difference of  across each of them because voltage is not divided in parallel combination.

Let $R_{1} , R_{2} , R_{3}$ be $10$Ω, $20$Ω and $60$Ω.

Let $R_{T}$ be the total resistance.

$\frac{1}{R_{T} } =\frac{1}{R_{1} } +\frac{1}{R_{2} }+\frac{1}{R_{3} }\\\frac{1}{R_{T} }=\frac{1}{10} +\frac{1}{20} +\frac{1}{60} \\\frac{1}{R_{T} }=\frac{6+3+1}{60} =\frac{10}{60} \\R_{T} =6$

Answer:

So, Total effective resistance is $6$Ω.