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Current Electricity

268428 Twelve resistances each of resistance $R$ are connected in the circuit as shown in fig. Net resistance between points $A$ and $C$ would be

1

\frac{5 R}{3}

2

\frac{7 R}{6}

3

R

4

\frac{3 R}{4}

Explanation:

When a cell is connected across $A$ and $C$ , no current flows in the arms BG and ED due to symmetry in the arrangement. Then equivalent circuit will be as shown in Fig. (a) and (b). The effective resistance between $A$ and $C$ is
$\begin{aligned} \frac{1}{R_{e f f}} = \frac{1}{2 R} + \frac{1}{3 R} + \frac{1}{2 R} \\ = \frac{3 + 2 + 3}{6 R} = \frac{8}{6 R} = \frac{4}{3 R} or R_{e f f} = \frac{3 R}{4} \end{aligned}$

(a)

Current Electricity

268429 A resistance is made by connecting two wires (series) of same material of radii $2 mm$ and 5 $mm$ and length $8 cm$ and $5 cm$ . A potential difference of $22 V$ is applied to them. The potential difference on the longer wire is

1

15 V

2

18 V

3

16 V

4

20 V

Current Electricity

268430 A $220 V$ and $800 W$ electric kettle and three $220 V$ and $100 W$ bulbs are connected in parallel. O n connecting this combination with $200 V$ supply, the total current in the circuit will be

1

0.15 A

2

5 A

3

5.5 A

4

4.55 A

Explanation:

$R = \frac{V^{2}}{P}$ ; find $R_{e q}$ then $i = \frac{V}{R_{e q}}$

Current Electricity

268431 What is the equivalent resistance of the circuit

1

6 Ω

2

7 Ω

3

8 Ω

4

9 Ω

Current Electricity

268428 Twelve resistances each of resistance $R$ are connected in the circuit as shown in fig. Net resistance between points $A$ and $C$ would be

1

\frac{5 R}{3}

2

\frac{7 R}{6}

3

R

4

\frac{3 R}{4}

Explanation:

When a cell is connected across $A$ and $C$ , no current flows in the arms BG and ED due to symmetry in the arrangement. Then equivalent circuit will be as shown in Fig. (a) and (b). The effective resistance between $A$ and $C$ is
$\begin{aligned} \frac{1}{R_{e f f}} = \frac{1}{2 R} + \frac{1}{3 R} + \frac{1}{2 R} \\ = \frac{3 + 2 + 3}{6 R} = \frac{8}{6 R} = \frac{4}{3 R} or R_{e f f} = \frac{3 R}{4} \end{aligned}$

(a)

Current Electricity

268429 A resistance is made by connecting two wires (series) of same material of radii $2 mm$ and 5 $mm$ and length $8 cm$ and $5 cm$ . A potential difference of $22 V$ is applied to them. The potential difference on the longer wire is

1

15 V

2

18 V

3

16 V

4

20 V

Current Electricity

268430 A $220 V$ and $800 W$ electric kettle and three $220 V$ and $100 W$ bulbs are connected in parallel. O n connecting this combination with $200 V$ supply, the total current in the circuit will be

1

0.15 A

2

5 A

3

5.5 A

4

4.55 A

Explanation:

$R = \frac{V^{2}}{P}$ ; find $R_{e q}$ then $i = \frac{V}{R_{e q}}$

Current Electricity

268431 What is the equivalent resistance of the circuit

1

6 Ω

2

7 Ω

3

8 Ω

4

9 Ω

Current Electricity

268428 Twelve resistances each of resistance $R$ are connected in the circuit as shown in fig. Net resistance between points $A$ and $C$ would be

1

\frac{5 R}{3}

2

\frac{7 R}{6}

3

R

4

\frac{3 R}{4}

Explanation:

When a cell is connected across $A$ and $C$ , no current flows in the arms BG and ED due to symmetry in the arrangement. Then equivalent circuit will be as shown in Fig. (a) and (b). The effective resistance between $A$ and $C$ is
$\begin{aligned} \frac{1}{R_{e f f}} = \frac{1}{2 R} + \frac{1}{3 R} + \frac{1}{2 R} \\ = \frac{3 + 2 + 3}{6 R} = \frac{8}{6 R} = \frac{4}{3 R} or R_{e f f} = \frac{3 R}{4} \end{aligned}$

(a)

Current Electricity

268429 A resistance is made by connecting two wires (series) of same material of radii $2 mm$ and 5 $mm$ and length $8 cm$ and $5 cm$ . A potential difference of $22 V$ is applied to them. The potential difference on the longer wire is

1

15 V

2

18 V

3

16 V

4

20 V

Current Electricity

268430 A $220 V$ and $800 W$ electric kettle and three $220 V$ and $100 W$ bulbs are connected in parallel. O n connecting this combination with $200 V$ supply, the total current in the circuit will be

1

0.15 A

2

5 A

3

5.5 A

4

4.55 A

Explanation:

$R = \frac{V^{2}}{P}$ ; find $R_{e q}$ then $i = \frac{V}{R_{e q}}$

Current Electricity

268431 What is the equivalent resistance of the circuit

1

6 Ω

2

7 Ω

3

8 Ω

4

9 Ω

Current Electricity

268428 Twelve resistances each of resistance $R$ are connected in the circuit as shown in fig. Net resistance between points $A$ and $C$ would be

1

\frac{5 R}{3}

2

\frac{7 R}{6}

3

R

4

\frac{3 R}{4}

Explanation:

When a cell is connected across $A$ and $C$ , no current flows in the arms BG and ED due to symmetry in the arrangement. Then equivalent circuit will be as shown in Fig. (a) and (b). The effective resistance between $A$ and $C$ is
$\begin{aligned} \frac{1}{R_{e f f}} = \frac{1}{2 R} + \frac{1}{3 R} + \frac{1}{2 R} \\ = \frac{3 + 2 + 3}{6 R} = \frac{8}{6 R} = \frac{4}{3 R} or R_{e f f} = \frac{3 R}{4} \end{aligned}$

(a)

Current Electricity

268429 A resistance is made by connecting two wires (series) of same material of radii $2 mm$ and 5 $mm$ and length $8 cm$ and $5 cm$ . A potential difference of $22 V$ is applied to them. The potential difference on the longer wire is

1

15 V

2

18 V

3

16 V

4

20 V

Current Electricity

268430 A $220 V$ and $800 W$ electric kettle and three $220 V$ and $100 W$ bulbs are connected in parallel. O n connecting this combination with $200 V$ supply, the total current in the circuit will be

1

0.15 A

2

5 A

3

5.5 A

4

4.55 A

Explanation:

$R = \frac{V^{2}}{P}$ ; find $R_{e q}$ then $i = \frac{V}{R_{e q}}$

Current Electricity

268431 What is the equivalent resistance of the circuit

1

6 Ω

2

7 Ω

3

8 Ω

4

9 Ω