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PHXII03:CURRENT ELECTRICITY

357346 Find the current in $6 Ω$ resistance in the circuit show in figure.
supporting img

1

1 A

2

5 A

3

9 A

4

3 A

Explanation:

The Wheatstone bridge in given circuit is balanced, we can remove its middle branch. After this remaining resistance of $16 Ω, 8 Ω$ and $16 Ω$ will be in parallel for which we can find the equivalent resistance, $R_{e q} = {(\frac{1}{16} + \frac{1}{8} + \frac{1}{16})}^{- 1} + 1 = 5 Ω$
Current supplied by the battery, $I = \frac{V}{R_{e q}} = \frac{60}{5} = 12 A$
In the circuit 12 A will be distributed in the three branches in parallel. The potential drop across the three branches in parallel, $X Y = I R_{X Y} = 12 \times 4 = 48 V$
Thus, current in top $16 Ω$ resistance branch, $I_{6 Ω} = \frac{V_{K Y}}{16} = \frac{48}{16} = 3 A$

PHXII03:CURRENT ELECTRICITY

357347 The equivalent resistance across $A$ and $B$ is
supporting img

1

2 Ω

2

3 Ω

3

4 Ω

4

5 Ω

Explanation:

The equivalent circuit can be redrawn as
supporting img
We have $\frac{P}{Q} = \frac{R}{S}$ , i.e. $\frac{4}{4} = \frac{4}{4}$ So, the given circuit is a balanced Wheatstone bridge. Hence, the equivalent resistance
$R_{A B} = \frac{(4 + 4) \times (4 + 4)}{(4 + 4) + (4 + 4)}$
$= \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 Ω$

PHXII03:CURRENT ELECTRICITY

357348 The current $I$ drawn from the $5 V$ source will be
supporting img

1

0.33 A

2

0.5 A

3

0.67 A

4

0.17 A

Explanation:

The given circuit can be redrawn as
supporting img
The circuit shows the balanced condition of Wheatstone bridge and hence no current flows in the middle resistor. So, equivalent circuit would redrawn be as shown below

$10 Ω$ and $20 Ω$ resistances are in series, then $R^{'} = 10 Ω + 20 Ω = 30 Ω$
Similarly, $5 Ω$ and $10 Ω$ are in series, then $R^{''} = 15 Ω$
Now, $R^{'}$ and $R^{''}$ are in parallel, then
$R = \frac{15 \times 30}{15 + 30} = 10 Ω$
So, $I = \frac{V}{R} = \frac{5}{10} = 0.5 A$

PHXII03:CURRENT ELECTRICITY

357349 In the following circuit, the reading of the voltmeter will be: (in volt)
supporting img

1 7.2

2 4.8

3 6

4 4

Explanation:

Voltage on $40 cm = 4.8 V$
Voltage on $60 cm$ so $= \frac{4.5}{40} \times 60 = 7.2$ volt .

PHXII03:CURRENT ELECTRICITY

357346 Find the current in $6 Ω$ resistance in the circuit show in figure.
supporting img

1

1 A

2

5 A

3

9 A

4

3 A

Explanation:

The Wheatstone bridge in given circuit is balanced, we can remove its middle branch. After this remaining resistance of $16 Ω, 8 Ω$ and $16 Ω$ will be in parallel for which we can find the equivalent resistance, $R_{e q} = {(\frac{1}{16} + \frac{1}{8} + \frac{1}{16})}^{- 1} + 1 = 5 Ω$
Current supplied by the battery, $I = \frac{V}{R_{e q}} = \frac{60}{5} = 12 A$
In the circuit 12 A will be distributed in the three branches in parallel. The potential drop across the three branches in parallel, $X Y = I R_{X Y} = 12 \times 4 = 48 V$
Thus, current in top $16 Ω$ resistance branch, $I_{6 Ω} = \frac{V_{K Y}}{16} = \frac{48}{16} = 3 A$

PHXII03:CURRENT ELECTRICITY

357347 The equivalent resistance across $A$ and $B$ is
supporting img

1

2 Ω

2

3 Ω

3

4 Ω

4

5 Ω

Explanation:

The equivalent circuit can be redrawn as
supporting img
We have $\frac{P}{Q} = \frac{R}{S}$ , i.e. $\frac{4}{4} = \frac{4}{4}$ So, the given circuit is a balanced Wheatstone bridge. Hence, the equivalent resistance
$R_{A B} = \frac{(4 + 4) \times (4 + 4)}{(4 + 4) + (4 + 4)}$
$= \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 Ω$

PHXII03:CURRENT ELECTRICITY

357348 The current $I$ drawn from the $5 V$ source will be
supporting img

1

0.33 A

2

0.5 A

3

0.67 A

4

0.17 A

Explanation:

The given circuit can be redrawn as
supporting img
The circuit shows the balanced condition of Wheatstone bridge and hence no current flows in the middle resistor. So, equivalent circuit would redrawn be as shown below

$10 Ω$ and $20 Ω$ resistances are in series, then $R^{'} = 10 Ω + 20 Ω = 30 Ω$
Similarly, $5 Ω$ and $10 Ω$ are in series, then $R^{''} = 15 Ω$
Now, $R^{'}$ and $R^{''}$ are in parallel, then
$R = \frac{15 \times 30}{15 + 30} = 10 Ω$
So, $I = \frac{V}{R} = \frac{5}{10} = 0.5 A$

PHXII03:CURRENT ELECTRICITY

357349 In the following circuit, the reading of the voltmeter will be: (in volt)
supporting img

1 7.2

2 4.8

3 6

4 4

Explanation:

Voltage on $40 cm = 4.8 V$
Voltage on $60 cm$ so $= \frac{4.5}{40} \times 60 = 7.2$ volt .

PHXII03:CURRENT ELECTRICITY

357346 Find the current in $6 Ω$ resistance in the circuit show in figure.
supporting img

1

1 A

2

5 A

3

9 A

4

3 A

Explanation:

The Wheatstone bridge in given circuit is balanced, we can remove its middle branch. After this remaining resistance of $16 Ω, 8 Ω$ and $16 Ω$ will be in parallel for which we can find the equivalent resistance, $R_{e q} = {(\frac{1}{16} + \frac{1}{8} + \frac{1}{16})}^{- 1} + 1 = 5 Ω$
Current supplied by the battery, $I = \frac{V}{R_{e q}} = \frac{60}{5} = 12 A$
In the circuit 12 A will be distributed in the three branches in parallel. The potential drop across the three branches in parallel, $X Y = I R_{X Y} = 12 \times 4 = 48 V$
Thus, current in top $16 Ω$ resistance branch, $I_{6 Ω} = \frac{V_{K Y}}{16} = \frac{48}{16} = 3 A$

PHXII03:CURRENT ELECTRICITY

357347 The equivalent resistance across $A$ and $B$ is
supporting img

1

2 Ω

2

3 Ω

3

4 Ω

4

5 Ω

Explanation:

The equivalent circuit can be redrawn as
supporting img
We have $\frac{P}{Q} = \frac{R}{S}$ , i.e. $\frac{4}{4} = \frac{4}{4}$ So, the given circuit is a balanced Wheatstone bridge. Hence, the equivalent resistance
$R_{A B} = \frac{(4 + 4) \times (4 + 4)}{(4 + 4) + (4 + 4)}$
$= \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 Ω$

PHXII03:CURRENT ELECTRICITY

357348 The current $I$ drawn from the $5 V$ source will be
supporting img

1

0.33 A

2

0.5 A

3

0.67 A

4

0.17 A

Explanation:

The given circuit can be redrawn as
supporting img
The circuit shows the balanced condition of Wheatstone bridge and hence no current flows in the middle resistor. So, equivalent circuit would redrawn be as shown below

$10 Ω$ and $20 Ω$ resistances are in series, then $R^{'} = 10 Ω + 20 Ω = 30 Ω$
Similarly, $5 Ω$ and $10 Ω$ are in series, then $R^{''} = 15 Ω$
Now, $R^{'}$ and $R^{''}$ are in parallel, then
$R = \frac{15 \times 30}{15 + 30} = 10 Ω$
So, $I = \frac{V}{R} = \frac{5}{10} = 0.5 A$

PHXII03:CURRENT ELECTRICITY

357349 In the following circuit, the reading of the voltmeter will be: (in volt)
supporting img

1 7.2

2 4.8

3 6

4 4

Explanation:

Voltage on $40 cm = 4.8 V$
Voltage on $60 cm$ so $= \frac{4.5}{40} \times 60 = 7.2$ volt .

PHXII03:CURRENT ELECTRICITY

357346 Find the current in $6 Ω$ resistance in the circuit show in figure.
supporting img

1

1 A

2

5 A

3

9 A

4

3 A

Explanation:

The Wheatstone bridge in given circuit is balanced, we can remove its middle branch. After this remaining resistance of $16 Ω, 8 Ω$ and $16 Ω$ will be in parallel for which we can find the equivalent resistance, $R_{e q} = {(\frac{1}{16} + \frac{1}{8} + \frac{1}{16})}^{- 1} + 1 = 5 Ω$
Current supplied by the battery, $I = \frac{V}{R_{e q}} = \frac{60}{5} = 12 A$
In the circuit 12 A will be distributed in the three branches in parallel. The potential drop across the three branches in parallel, $X Y = I R_{X Y} = 12 \times 4 = 48 V$
Thus, current in top $16 Ω$ resistance branch, $I_{6 Ω} = \frac{V_{K Y}}{16} = \frac{48}{16} = 3 A$

PHXII03:CURRENT ELECTRICITY

357347 The equivalent resistance across $A$ and $B$ is
supporting img

1

2 Ω

2

3 Ω

3

4 Ω

4

5 Ω

Explanation:

The equivalent circuit can be redrawn as
supporting img
We have $\frac{P}{Q} = \frac{R}{S}$ , i.e. $\frac{4}{4} = \frac{4}{4}$ So, the given circuit is a balanced Wheatstone bridge. Hence, the equivalent resistance
$R_{A B} = \frac{(4 + 4) \times (4 + 4)}{(4 + 4) + (4 + 4)}$
$= \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 Ω$

PHXII03:CURRENT ELECTRICITY

357348 The current $I$ drawn from the $5 V$ source will be
supporting img

1

0.33 A

2

0.5 A

3

0.67 A

4

0.17 A

Explanation:

The given circuit can be redrawn as
supporting img
The circuit shows the balanced condition of Wheatstone bridge and hence no current flows in the middle resistor. So, equivalent circuit would redrawn be as shown below

$10 Ω$ and $20 Ω$ resistances are in series, then $R^{'} = 10 Ω + 20 Ω = 30 Ω$
Similarly, $5 Ω$ and $10 Ω$ are in series, then $R^{''} = 15 Ω$
Now, $R^{'}$ and $R^{''}$ are in parallel, then
$R = \frac{15 \times 30}{15 + 30} = 10 Ω$
So, $I = \frac{V}{R} = \frac{5}{10} = 0.5 A$

PHXII03:CURRENT ELECTRICITY

357349 In the following circuit, the reading of the voltmeter will be: (in volt)
supporting img

1 7.2

2 4.8

3 6

4 4

Explanation:

Voltage on $40 cm = 4.8 V$
Voltage on $60 cm$ so $= \frac{4.5}{40} \times 60 = 7.2$ volt .

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