Question

Compute the heat generated while transferring 72000 coulomb of charge in one hour through a potential difference of 50 V. [2M]

Answer 1

Explanation:

CURRENT (I) = Q/t

= 72000 / 60*60 (time in second)

= 20 ampere

Potential difference (V) = 50V

Time (t) = 60*60 = 3600s

Heat generated = VIt = 50*20*3600

= 36,00,000J

= 36*10^5J

Alternative method :

Heat generated = VIt = V*Q/t *t

= V*Q

= 72000*50

=36,00,000

=3.6*10^5J

Answer 2

GIVEN :-

• Charge , Q = 72000 coulomb

• Time , t = 60 × 60 seconds

• Voltage , V = 50 V

TO FIND :-

• Amount of heat generated.

SOLUTION :-

 We know that ,

      Q = It

 Here, I = current

 $\longrightarrow\sf I=\dfrac{Q}{t} \\\\\longrightarrow I = \dfrac{72000}{60\times 60 } \\\\\longrightarrow I = \dfrac{72000}{3600} \\\\\longrightarrow \bf I = 20 \ A$

Formula to find heat ,

 H = VIt

 $\longrightarrow\sf H = 50 \times 20 \times 60 \times 60 \\\\\longrightarrow\bf H = 3600000 \ J$

ALTERNATIVE METHOD

Given ,

Voltage , V = 50 V

Current , I = 20 A

We know that ,

  V = IR

Here , R = Resistance

$\longrightarrow\sf R = \dfrac{V}{I}\\\\\longrightarrow R = \dfrac{50}{20}\\\\\longrightarrow \bf R =\dfrac{5}{2} \ \Omega$

Formula to find heat ,

 H = I²Rt

 $\longrightarrow\sf H = 20 \times 20 \times \dfrac{5}{2} \times 60 \times 60 \\\\\longrightarrow H = 20 \times 20 \times 5 \times 30 \times 60 \\\\\longrightarrow \bf H= 3600000 \ J$