Question

200j of heat is produced 10 sec in a 5 resistance find the potential difference across the resistor

Answer 1

Answer:

• Potential difference = 10 V

Explanation

Given:-

• Heat produced, H = 200 J
• Time, t = 10 sec
• Resistance, R = 5 Ω

To find:-

• Potential difference across the resistor, V =?

Formula required:-

• Joule's equation of electrical heating

        H = I²Rt

• Ohm's law expression

      V = IR

[ Where H is heat produced, I is current, R is effective resistance, t is time and V is potential difference ]

Solution:-

Using the Joules's equation of heating

→ H = I²RT

[ by the Ohm's law ]

→ H = (V/R)²RT

→ H = (V²/R²) RT  

→ H = V²T/R

[ putting values ]

→ 200 = V² × 10 / 5

→ 200 = 2 V²

→ V² = 100

→ V = 10 V

Therefore,

The Potential difference across the resistor is 10 Volts.

Answer 2

Explanation:

$\frak Given = \begin{cases} &\sf{Heat\ produced,\ H\ =\ 200\ J.} \\ &\sf{Time,\ t\ =\ 10\ sec.} \\ &\sf{Resistance,\ R\ =\ 5\ Ω.} \end{cases}$

To find:- We have to find the Potential difference across the resistor, V = ?

__________________

$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf\pink{\underline{Joule's\ equation\ of\ electrical\ heating:-}}$

$\sf : \implies {H\ =\ I²Rt}$

$\sf\pink{\underline{Ohm's\ law\ expression:-}}$

$\sf : \implies {V\ =\ IR}$

Here:-

● H is for heat produced.

● I is for current.

● R is for effective resistance.

● t is for time.

● V is for potential difference.

__________________

$\frak{\underline{\underline{\dag By\ Joules\ equation\ of\ heating:-}}}$

$\sf : \implies {H\ =\ I²Rt}$

$\frak{\underline{\underline{\dag By\ Ohm's\ law\ of\ expression:-}}}$

$\sf : \implies {H\ =\ \bigg (\dfrac{V}{R} \bigg)^2\ Rt} \\ \\ \sf : \implies {H\ =\ \bigg (\dfrac{V²}{R²} \bigg)\ Rt} \\ \\ \sf : \implies {H\ =\ \dfrac{V²t}{R}}$

__________________

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {200\ =\ \dfrac{V²×10}{5}} \\ \\ \sf : \implies {200\ =\ 2V²} \\ \\ \sf : \implies {V²\ =\ 100} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf V\ =\ 10\ V.}}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ potential\ difference\ across\ the\ resistor\ is\ 10\ V.}}$