Question

A one-ohm and half-ohm resistor are connected in parallel across a 3 volt battery. Total energy given out per second is

Answer 1

Answer:

$\boxed{\sf Total \ energy \ given \ out \ per \ second = 27 \ J}$

Given:

Value of resistance connected in series combination:

$\sf R_1 = 1 \Omega$

$\sf R_2 = \frac{1}{2} \Omega$

Potential difference of battery (V) = 3 Volt

To Find:

Total energy given out per second in the circuit

Explanation:

Effective resistance:

$\boxed{ \bold{\frac{1}{R_{eff}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}}}$

Substituting values of $\sf R_1 \ and \ R_2$ in the equation:

$\sf \implies \frac{1}{R_{eff}} = \frac{1}{1} + \frac{1}{ \frac{1}{2} }$

$\sf \implies \frac{1}{R_{eff}} = 1 + 2$

$\sf \implies \frac{1}{R_{eff}} = 3$

$\sf \implies R_{eff} = \frac{1}{3} \Omega$

Electrical energy formula:

$\boxed{ \bold{\sf E = \frac{V^2}{R_{eff}}t}}$

Here, t = 1 second as we need to calculate total energy given out per second.

Substituting values of V, $\sf R_{eff}$ & t in the equation:

$\sf \implies E = \frac{ {3}^{2} }{ \frac{1}{3} } \times 1$

$\sf \implies E = {3}^{2} \times 3$

$\sf \implies E = {3}^{3}$

$\sf \implies E = 27 \: J$

$\therefore$

Total energy given out per second in the circuit = 27 Joule

Answer 2

Explanation:

27 J

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