# Current Electricity - Class 12 - Physics

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## Extra Questions - Current Electricity | NCERT | Physics | Class 12

What effect does adding lightbulbs have in a series circuit? Describe how the voltage changed across the different points in a series circuit.

When additional lightbulbs are added to a series circuit, several key changes occur, especially in terms of voltage distribution and power dissipation across each bulb, which in turn affects brightness.

**Voltage Distribution Across Bulbs:**In a series circuit, the total voltage supplied by the source is divided equally among all the bulbs assuming they have equal resistance. Consider a scenario with three bulbs connected in series to a voltage source $ V $:

If each bulb has resistance $ R $, the total resistance $ R_{\text{eq}} $ of the circuit is the sum of the resistances of each bulb. So, $ R_{\text{eq}} = R + R + R = 3R $.

The current $ I $ flowing through the circuit can be expressed as: $$ I = \frac{V}{R_{\text{eq}}} = \frac{V}{3R} $$

Voltage across each bulb then becomes: $$ V_{\text{AB}} = V_{\text{BC}} = V_{\text{CD}} = I \times R = \frac{V}{3} $$

If an additional bulb is added, making it four bulbs in series: $$ R_{\text{eq}} = R + R + R + R = 4R $$

The new current $ I' $ would be: $$ I' = \frac{V}{4R} $$

Voltage across each bulb now: $$ V'

*{\text{AB}} = V'*{\text{BC}} = V'*{\text{CD}} = V'*{\text{DE}} = I' \times R = \frac{V}{4} $$

**Power Dissipation and Brightness:**The power dissipated in each bulb, which relates to its brightness, is given by $ P = I^2 \times R $.

Initially with three bulbs: $$ P = \left(\frac{V}{3R}\right)^2 \times R = \frac{V^2}{9R} $$

After adding the fourth bulb: $$ P' = \left(\frac{V}{4R}\right)^2 \times R = \frac{V^2}{16R} $$

**Significant Observation:**The power dissipation per bulb $ P' $ is less than $ P $. Hence,**the brightness of each bulb decreases as more bulbs are added to the series.**

In summary, adding bulbs to a series circuit results in a reduction of both the voltage drop across each bulb and the brightness of each bulb due to lower power dissipation. The overall resistance of the circuit increases, which in turn reduces the current flowing through each bulb, consequently decreasing the voltage across and the power dissipated by each bulb.

Heating in a current-carrying conductor is a product of _____ and time.

A. current

B. resistance

C. power

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What is the working principle of an electrical fuse?

A) Ohm's law

B) Joule's law of heating

C) Coulomb's law

D) Faraday's law

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If the resistance of a conductor is kept constant and the voltage through the appliance is reduced to half. Then, the resultant power is

(A) $\frac{1}{2} \times P$

(B) $\frac{1}{4} \times P$

(C) $4 \times P$

(D) $\frac{1}{8} \times P$

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Eleven $15 \Omega$ and four $7 \Omega$ resistors are connected in series. If a new resistance of $21 \Omega$ is added in series, the change in the effective resistance would be:

A) $21 \Omega$

B) $193 \Omega$

C) $234 \Omega$

D) $15 \Omega"

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"A resistance of $10 \Omega$ is bent to form a closed circle. What is the effective resistance between the two ends of any diameter of this circuit?"