Question

A battery of e.m.f. 15 V and internal resistance 2 $\Omega$ is connected to two resistors of resistance 4 ohm and 6 ohm joined in series. What is the electrical energy spent per minute in 6 ohm resistor ?

• A. 562.5 J
• B. 180 J
• C. 456.5 J
• D. 360 J

Answer 1

The correct option is A 562.5 J

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.
The total resistance of resistors in series is equal to the sum of their individual resistances. Hence, the Equivalent resistance is more than the individual resistances because a sum is taken of all the individual resistances. That is, $R_{total}=R_1+R_2$. The current is given as $I=I_1=I_2$.
In this case, the combined resistance of the 4 ohms and 6 ohms resistors is given as 4+6 = 10 ohms.
The total current in the circuit is given as $I=\frac{\epsilon }{R+r}=\frac{15}{10+2}=1.25A$
As it is a serially connected circuit the current remains the same and hence the potential difference is calculated from V=IR.
That is, $V=IR=1.25\times 6=7.5V$.
Energy ($power\times time$) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.
Energy dissipated = Pt or $VI\times tasP=VI$.
It is determined that the current I is 1.25 A, voltage V is 7.5 V and the time taken is 1 minute, that is, $1\times 60=60seconds$. The energy generated is given as $Q=VI\times t$.
Hence, the total energy by the 6 ohms resistor in 1 minute is $1.25A\times 7.5V\times 60s=562.5Joules$.