Question

An electric bulb of resistance 20 and a resistance wire of 4 are connected in series with a 6 V battery.draw the circuit diagram and calculate.(i)the total resistance of the circuit.(ii)current through the circuit.(iii potential differences across the electrical bulb.(iv)potential differences across the resistance wire.

Answer 1

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Given:

Resistance of bulb = 20 ohm

Resistance wire = 4 ohm

Voltage of battery = 6V

To find:

• Total resistance
• Total current
• Potential Difference along Bulb
• Potential Difference along wire

Diagram:

See the attached photo to understand better.

Calculation:

[From Ohm's Law, we know the basic relationship between Current (I) , resistance (R) and Voltage (V) :

V = I × R ]

The resistors are present in series connection.

1. Total resistance = 4 + 20 = 24 ohms.

2. Total current in circuit = 6/24 =0.25 A

3. Potential Difference across electric bulb

= (total current × resistance of bulb)

= 0.25 × 20

= 5 V

4. Potential Difference along resistance wire

= (total current × resistance of wire)

= 0.25 × 4

= 1 V

Additional information:

• Current along resistances in series connection remains equal.
• Potential Difference along resistance in parallel connection remains equal
• Voltage gets divided in series, whereas the current gets divided in parallel connection.

Answer 2

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Given :

• Resistance of Bulb (R1)= 20 ohm
• Resistance of Wire (R2) = 4 ohm
• Potential Difference (V) = 6V

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To Find :

• Total Resistance
• Current through circuit
• Potential difference across bulb
• Potential difference across wire

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Solution :

We have formula for Resistance in series :

$\large{\boxed{\sf{R_s \: = \: R_1 \: + \: R_2}}} \\ \\ \implies {\sf{R_s \: = \: 20 \: + \: 4}}$

$\therefore$ Resistance is 24 ohm

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Now, use formula for Current

$\large{\boxed{\sf{V \: = \: IR}}} \\ \\ \implies {\sf{I \: = \: \dfrac{V}{R}}} \\ \\ \implies {\sf{I \: = \: \dfrac{6}{24}}} \\ \\ \implies {\sf{I \: = \: 0.25}}$

Current in circuit is 0.25V

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Potential Difference across Bulb is 5 V

$\implies {\sf{V \: = \: IR}} \\ \\ \implies {\sf{V \: = \: 0.25 \: \times \: 20}} \\ \\ \implies {\sf{V \: = \: 5}}$

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Potential Difference across wire is 1V

$\implies {\sf{V \: = \: IR}} \\ \\ \implies {\sf{V \: = \: 0.25 \: \times \: 4}} \\ \\ \implies {\sf{V \: = \: 1}}$