Question

Figure shows a potential divider circuit, which by adjustment of the position of the contact X, can be used to provide a variable potential difference between the terminals P and Q. What are the limits of this potential difference?

A. $$0$$ and $$20\text{mV}$$
B. $$\text{5}$$ and $$25\text{mV}$$
C. $$0$$ and $$20\text{V}$$
D. $$0$$ and $$25\text{V}$$

Answer 1

Hint: We are asked to find the lower limit and upper limit of potential difference between the terminals P and Q. To find the lower limit move the terminal P to the lower end of resistor $$\text{4}\text{k}\mathrm{\Omega }$$ and to find the upper limit, move the terminal P to the upper end of resistor $$\text{4}\text{k}\mathrm{\Omega }$$.

Complete step by step answer:
We are given a circuit and asked to find the range of potential difference between the terminals P and Q.Let us first find the lower limit of the potential difference between the terminals P and Q.We can find the lower limit of potential difference by moving the terminal P to the lower end of $$\text{4}\text{k}\mathrm{\Omega }$$ resistor that is towards terminal Q. In this case, the potential difference between P and Q will be zero. Therefore, the lower limit of the potential difference between the terminals P and Q will be $$0\text{V}$$.

Now, we find the upper limit of potential difference between the terminals P and Q. In this case we move the terminal P to the upper end of $$\text{4}\text{k}\mathrm{\Omega }$$ resistor that is away from the terminal Q.Here, let $$i$$ be the current flowing through the circuit.
We redraw the circuit.

Here, the equivalent resistance will be,
$$R=\text{4}\text{k}\mathrm{\Omega }+1\text{kOmega }$$
$$\Rightarrow R=5\text{k}\mathrm{\Omega }$$
Current through a circuit is,
$$\text{voltage}=\text{current}\times \text{resistance}$$ (i)
Here, voltage is $$25\text{V}$$
Putting the values of voltage and equivalent resistance in equation (i) we get,
$$25=i\times 5$$
$$\Rightarrow i={\displaystyle \frac{25}{5}}=5\text{A}$$
Now, voltage across the resistor $$\text{4}\text{k}\mathrm{\Omega }$$ will be (using equation (i))
$$V_{PQ}=i\times 4$$
Putting the value of $$i$$ in the above equation we get,
$$V_{PQ}=5\times 4$$
$$\therefore V_{PQ}=20\text{V}$$
Therefore, the upper limit of potential difference between the terminals P and Q is $$20\text{V}$$.So, the limits of potential difference between the terminals P and Q are $$0\text{V}$$and $$20\text{V}$$.

Hence, the correct answer is option C.

Note:Using a potential divider circuit, we can get different voltages from a common supply voltage. One example of a potential divider is potentiometer. Potential divider circuits can be used for controlling audio volume, for controlling temperature in the freezer.