Question

A 2V cell is connected to a 10 Ω resistor. How many electrons come out of the negative terminal of the cell in 2 minutes?​

Answer 1

Given that;

A 2V cell is connected to a 10Ω resistor.

• We are asked to find number of electrons come out of the negative terminal of the cell in 2 minutes (120 seconds).

First of all we need to find magnitude of current flowing through the circuit.

★ As per ohm's law, current flowing through a conductor is directly proportional to the applied potential difference.

Mathematically, V ∝ I

• R is the proportionality constant which denotes resistance$.$

• V = I × R

By substituting the given values;

→ 2 = I × 10

→ I = 2/10

→ I = 0.2 A

We know that;

• I = Q / t = n e / t

• n denotes number of electrons

• e denotes charge of an electron

→ 0.2 = (n × 1.6 × 10‾¹⁹) / 120

→ n × 1.6 × 10‾¹⁹ = 0.2 × 120

→ n = (24/1.6) × 10¹⁹

→ n = 15 × 10¹⁹ e‾ per second

Answer 2

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$\underline{\underline{\bf{Solution : -}}}$

From ohm's law We know that ;

$\: \: \: \: \: \: \: \: \: \: v \: = ir$

Where V is the voltage , I is the current and R is the Resistor.

Given : - v = 2 volts , R = 10 Ω

$\: i \: = \dfrac{v}{r}$

$\: \: i \: = \dfrac{2}{10}$

$i = 5a$

Also we know that ;

$i \: = \dfrac{ne}{t}$

Putting the Value of I in (i) We get,

$i \: = \dfrac{5 \: \times t}{e}$

$for \: e \: = 1.6 \times {10}^{ - 19} \: and C \:and \:t \: = 2 \: minute\: = \: 120\: seconds$

$n = \dfrac{2 × 120} {1.6 × 10¹⁹}$

$\implies{n = 15 × 10¹⁹ electrons}$

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