Question

Two resistors of resistances 3Ω and 6Ω are connected in series across a battery of 12V. The electrical energy consumed in one minute in 3Ω resistance is *





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Answer 1

Answer:

$\boxed{\sf Electrical \ energy \ consumed \ in \ 1 \ min \ in \ 3 \Omega \ resistance = 320 \ J}$

Given:

Value of resistance connected in series:

$\sf R_{1} = 3 \Omega$

$\sf R_{2} = 6 \Omega$

Potential difference of the battery across resistors (V) = 12 V

Time (t) = 1 min

To Find:

Electrical energy consumed in one minute in 3Ω resistance

Explanation:

As the resistance are connected in series, Effective resistance of the circuit will be:

$\sf R_{eff} = R_1 + R_2$

$\sf R_{eff} = 3 \Omega + 6 \Omega$

$\sf R_{eff} = 9 \Omega$

$\therefore$

$\sf I = \frac{V}{R}$

$\sf I = \frac{12}{9}$

$\sf I = \frac{4}{3}A$

• Formula of Electrical Energy (E) = I²Rt

t = 1 min = 60 sec

Electrical energy for 3Ω resistance:

$\sf \implies E = (\frac{4}{3})^{2} \times 3 \times 60$

$\sf \implies E = \frac{16}{9} \times 3 \times 60$

$\sf \implies E = \frac{16}{3} \times 60$

$\sf \implies E = 16 \times 20$

$\sf \implies E = 320 \: J$