Name: Introduction to Van de Graaff Generator
Uploaded: 2019-07-19
Duration: 5 min 52 s
Description: If you keep charging a metallic conductor, it will eventually become a potential bomb. A machine was specially designed to generate and handle such high pote...

Question

Blessy extends her investigation to include circuit $b$ . She also decides to measure the current in each lamp and the cell. Here is a partly completed table of results.

Use your knowledge of current, voltage and also series and parallel circuits to complete the table of results with the values you would expect to observe.

Circuit	Current in amperes			Voltage in volts
	Cell	Lamp $1$	Lamp $2$	Cell	Lamp $1$	Lamp $2$
B	$3.0$	$3.0$		$1.5$	$0.75$
D	$3.0$	$1.5$

Accepted Answer

The given diagrams shows the two lamp are connected in series and parallel respectively.

Let $R$ be the resistance of each lamp.

Effective resistance in the series circuit is $R_{s e r i e s} = R_{1} + R_{2} + . . . . . . . + R_{n}$

Effective resistance in the parallel circuit is $R_{p a r a l l e l} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + . . . . . + \frac{1}{R_{n}}$

The total resistance in the circuit $B$ is $R_{s e r i e s} = 2 R$

The total resistance in the circuit $D$ is $R_{p a r a l l e l} = \frac{R}{2}$

We know that,

Current flowing in the circuit is $I = \frac{V}{R_{n e t}}$ (where $V$ is potential difference in the cell)

Given that, $V = 3.0 V$ , $I_{1} = 3.0 A$

$I = \frac{V}{R_{n e t}} \Rightarrow \frac{3.0}{2 R} = 3.0 \Rightarrow R = 0.5 Ω$

$I_{1} = \frac{V_{1}}{R_{1}} \Rightarrow \frac{V_{1}}{0.5} = 3.0 \Rightarrow V_{1} = 1.5$

The current flowing in the series circuit $B$ is same and the current in the parallel circuit splits depends on the resistance of the lamp.

Since the similar lamps are used in the parallel circuit $D$ the current flowing through the lamp $2$ is also $1.5 A$

If the cell is $1.5 V$ then, the potential difference across lamp $1$ is $V_{1} = 0.75 V$

Similarly the potential difference across lamp $2$ is also $V_{2} = 0.75 V$ $(∵ V = V_{1} + V_{2})$

In parallel circuit the potential difference $V = V_{1} = V_{2}$

When the cell $1.5 V$ is used in parallel circuit $D$ the potential difference across the cells are same as the cell.

Therefore, the table is given by

Circuit Current in amperes Voltage in volts

Cell Lamp $1$ Lamp $2$ Cell Lamp $1$ Lamp $2$

B $3.0$ $3.0$ $3.0$ $1.5$ $0.75$ $0.75$

D $3.0$ $1.5$ $1.5$ $1.5$ $1.5$ $1.5$

Length of resistance wire in cm	current in A
$10$	$2.0$
$20$	$2.2$
$30$	$0.8$
$40$	$1.2$

Length of resistance wire in cm	current in A
$10$	$2.0$
$20$	$2.2$
$30$	$0.8$
$40$	$1.2$

Important Questions on Electrostatics and Electric Currents

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