Question

in an electric circuit two resistor of 2 and 4 ohm repectively are connected in series to a 6 volt battert the heat dissipated by the forum register in 5 second will be​

Answer 1

Answer:

$\bigstar{\bold{Heat\:energy=30\:J}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Resistance (R₁) = 2 Ω
• Resistance (R₂) = 4 Ω
• Potential difference (V) = 6 V
• Time taken (t) = 5 s

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Heat energy dissipated

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

→ First we have to find the effective resistance of the circuit.

→ Since the resistors are connected in series, the effective resistance is given by

 R = R₁ + R₂

→ Substituting the data,

  R = 2 + 4

  R = 6 Ω

→ Hence the effective resistance of the circuit is 6 Ω

→ Now we have to find the current flowing in the circuit.

→ By Ohm's law we know that,

  V = I R

→ Substitute the data

  6 = I × 6

  I = 6/6

  I = 1 A

→ Hence the current flowing through the circuit is 1 Ampere.

→ Now the heat energy produced is given by Joule's law,

  H = I² R t

→ Substituting the data,

  H = 1² × 6 × 5

  H = 30 J

→ Hence the heat energy dissipated is 30 J

$\boxed{\bold{Heat\:energy=30\:J}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The effective resistance when resistors are connected in series is given by,

 $\sf{R=R_1+R_2+R_3+......+R_n}$

→ The effective resistance when resistors are connected in parallel is given by,

 $\sf{\dfrac{1}{R}=\dfrac{1}{R_1} +\dfrac{1}{R_2} +\dfrac{1}{R_3}+....+\dfrac{1}{R_n} }$

Answer 2

Resistance (R₁) = 2 Ω

Resistance (R₂) = 4 Ω

Potential difference (V) = 6 V

Time taken (t) = 5 s

Total resistance in series (R) = R₁+R₂

R = 2Ω + 4Ω

R = 6 Ω

$\boxed{ \rm{Heat ~Energy (h) = \frac{ (Potential difference)²×Time}{Resistance}}}$

$\large \sf h = \frac{6²×5}{6}$

$\large \sf h = \frac{36×5}{6}$

$\large \therefore \: \sf \underline{ \underline{h = 30J}}$