Question

3. Consider the following circuit
50
8
ww
12 0
w
OV​

Answer 1

Answer:

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Explanation:

From the given figure we ca easily find the resultant resistance of circuit,

⟹

R

eq

1

=

15

1

+

15

1

=

15

2

R

eq

=

2

15

Ω

Using formula for Ohm's law,

V=IR

eq

2=IR

eq

⟹I=

15

4

A

So, a current of

15

4

A is passing through circuit.

Using Kirchhoff's law for loop TRBSuT

⟹I

1

(10)+I

1

(5)−2=0

⟹I

1

=

15

2

A

⟹ Current through arm PAQ

⟹I−I

1

=

15

4

−

15

2

=

15

2

A

So, potential at point A is given by

V

A

=IR=

15

2

×5=

3

2

V

So, potential at point B is given by

V

B

=IR=

15

2

×10=

3

4

V

So, potential difference, V

A

−V

B

=

3

4

−

3

2

=

3

2

V

Answer 2

Answer:

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Explanation:

From the given figure we ca easily find the resultant resistance of circuit,

⟹

R

eq

1

=

15

1

+

15

1

=

15

2

R

eq

=

2

15

Ω

Using formula for Ohm's law,

V=IR

eq

2=IR

eq

⟹I=

15

4

A

So, a current of

15

4

A is passing through circuit.

Using Kirchhoff's law for loop TRBSuT

⟹I

1

(10)+I

1

(5)−2=0

⟹I

1

=

15

2

A

⟹ Current through arm PAQ

⟹I−I

1

=

15

4

−

15

2

=

15

2

A

So, potential at point A is given by

V

A

=IR=

15

2

×5=

3

2

V

So, potential at point B is given by

V

B

=IR=

15

2

×10=

3

4

V

So, potential difference, V

A

−V

B

=

3

4

−

3

2

=

3

2

V