A battery of emf and negligible internal resistance is connected to two resistances R1 and R2 in series. What is the potential difference across R2?(a)E(R1+R2)/R2(b)E(R1+R2)/R1(c)ER2/R1+R2(d)ER1/R1+R2

c. ER2/(R1+R2)

a)E(R1+R2)/R2

To determine the potential difference across R2, we need to apply Ohm's Law, which states that the potential difference (V) across a resistor is equal to the current (I) flowing through it multiplied by its resistance (R).

In this scenario, we are given a battery of electromotive force (emf) with negligible internal resistance, connected to resistances R1 and R2 in series. When resistors are connected in series, the current flowing through each resistor is the same.

Let's break down the problem and derive the expression for the potential difference across R2:

1. Determine the total resistance (RT) of the circuit:
Since R1 and R2 are in series, the total resistance (RT) is the sum of their individual resistances: RT = R1 + R2.

2. Find the current (I) flowing through the circuit:
Since the emf of the battery is connected to the resistances in series, the current flowing through R1 and R2 is the same. We can use Ohm's Law to express the current as I = E / RT, where E is the emf of the battery and RT is the total resistance.

3. Calculate the potential difference across R2 (VR2):
Using Ohm's Law, VR2 can be expressed as VR2 = I * R2. To obtain the value of I, plug in the expression I = E / RT. Thus, VR2 = (E / RT) * R2.

Now that we have derived the expression, let's simplify it to identify the correct option:

VR2 = (E / RT) * R2
= E * R2 / (R1 + R2)

Comparing the expression with the given options, we find that the correct answer is (c) ER2/(R1 + R2).

Therefore, the potential difference across R2 is ER2/(R1 + R2).