Question

The amount of work done in moving 5C charge is 25J. Calculate the
potential difference. What will be the resistance if the same potential
difference is applied across a cell connected a bulb and the current is
measured to be 4A.​

Answer 1

Answer :-

Given :-

• Work done = 5 C
• Charge = 25 J

To Find :-

• Potential difference
• Resistance

Solution :-

Potential difference = Work done / Charge

Substituting the values :-

→ V = 5 / 25

→ V = 1 / 5

→ V = 0.2

Potential difference = 0.2 Volt

Now, we have :-

• Potential difference = 0.2 V
• Current = 4 A

We know that,

Potential difference ( Voltage ) = Current × Resistance

Substituting the values :-

→ 0.2 = R × 4

→ R = 0.2 / 4

→ R = 0.05

Resistance = 0.05 Ω

Answer 2

Given :-

$~$

• Work done = 5 C
• Charge = 25 J

$~$

To find :-

$~$

• Potential difference
• Resistance

$~$

Solution :-

$~$

• Potential difference = $\large{\sf{\frac{Work \: done}{Charge} }}$

$~$

$~~~~~~~~~~~$★ Substituting the values :

$~$

:$\implies$ $\large{\sf{V = \frac{5}{25} }}$

$~$

$~~~~$:$\implies$ $\large{\sf{V = \frac{1}{5}}}$

$~$

$~~~~~~~~~~$:$\implies$ $\large{\underline{\boxed{\bf{\pink{V = 0.2}}}}}$★

$~$

• Potential difference = 0.2 Volt

$~$

Now,

$~$

• Potential difference = 0.2 V
• Current = 4 A

$~$

★ As we know that :

$~$

• Potential difference (Voltage) = Current × Resistance

$~$

★ Substituting the values,

$~$

:$\implies$ R × 4

$~$

$~~~~~$:$\implies$ $\large{\sf{\frac{0.2}{4}}}$

$~$

$~~~~~~~~~~$:$\implies$ $\large{\underline{\boxed{\bf{\purple{0.05}}}}}$★

$~$

$\large\dag$ Hence

$~$

• Resistance = 0.05 Ω