In Figure, circuit section ABabsorbs energy at a rate of 50 W when current i= 1.0 Although it is in the indicated direction. Resistance R= 2.0 Ω. (a) What is the potential difference between Aand B? Emf device Xlacks internal resistance.

(b) What is its emf? (c) Is point Bconnected to the positive terminal of Xor to the negative terminal?

Current i=1A

$$\begin{gathered} R=2\Omega \\ P=50W \end{gathered}$$

Use Ohm’s law to find the emf between the terminals and using equation of power, we can find the potential difference between the terminals.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Formulae are as follow:

$$\begin{gathered} V=iR \\ P=iV \end{gathered}$$

Where, i is current, V is voltage, R is resistance, P is power.

The potential difference between points A and B:

If i is the current flowing and $\Delta V$is the potential difference between points A and B then the power absorbed is given by,

$$\begin{gathered} P=i\Delta V \\ \Delta V=\frac{P}{i} \\ \Delta V=\frac{50W}{1A} \\ \Delta V=50V \end{gathered}$$

Since the energy of the charge decreases from point A to point B, the point A is at higher potential than point B.

The difference is $V_A-V_B=50V$

Hence, the potential difference between point A and B is 50V.

Emf of X :

The potential difference between point A and B is,

$\Delta V=V_A-V_B=iR+\epsilon$

Hence,

$$\begin{gathered} \epsilon =50V-iR \\ \epsilon =50V-(1A)(2\Omega ) \\ \epsilon =48V \end{gathered}$$

Hence, the Emf is 48V.

Since ε is positive and the current is toward right, B is connected to the negative terminal and A is to the positive terminal.

Hence, the point Bis connected at the negative terminal of X.

Therefore, using Ohm’s law, find the emf between the terminals. And using power equation, find the potential difference between the terminals.