Question

A potential divider circuit is connected with a dc source of 20 V, a light emitting diode of glow in voltage 1.8 V and a zener diode of breakdown voltage of 3.2 V. The length (PR) of the resistive wire is 20 cm. The minimum length of PQ to just glow the LED is …………. cm.

Answer 1

The total voltage across $PR$ is:
$V_{PR} = 20 \, \text{V}.$
The resistive wire $PR$ has a total length of:
$\ell_{PR} = 20 \, \text{cm}.$
The voltage across $QR$ is determined by the zener diode:
$V_{QR} = 3.2 \, \text{V}.$
The voltage across $PQ$ is:
$V_{PQ} = V_{PR} - V_{QR} = 20 - 3.2 = 16.8 \, \text{V}.$
The fraction of voltage across $PQ$ relative to $PR$ is:
$\frac{V_{PQ}}{V_{PR}} = \frac{16.8}{20} = \frac{1}{4}.$
Using the proportionality of voltage and length:
$\ell_{PQ} = \frac{1}{4} \times \ell_{PR}.$
Substitute $\ell_{PR} = 20 \, \text{cm}$:
$\ell_{PQ} = \frac{1}{4} \times 20 = 5 \, \text{cm}.$
Thus, the minimum length of $PQ$ to just glow the LED is:
$\ell_{PQ} = 5 \, \text{cm}.$

Answer 2

The total voltage across $PR$ is:
$V_{PR} = 20 \, \text{V}.$
The resistive wire $PR$ has a total length of:
$\ell_{PR} = 20 \, \text{cm}.$
The voltage across $QR$ is determined by the zener diode:
$V_{QR} = 3.2 \, \text{V}.$
The voltage across $PQ$ is:
$V_{PQ} = V_{PR} - V_{QR} = 20 - 3.2 = 16.8 \, \text{V}.$
The fraction of voltage across $PQ$ relative to $PR$ is:
$\frac{V_{PQ}}{V_{PR}} = \frac{16.8}{20} = \frac{1}{4}.$
Using the proportionality of voltage and length:
$\ell_{PQ} = \frac{1}{4} \times \ell_{PR}.$
Substitute $\ell_{PR} = 20 \, \text{cm}$:
$\ell_{PQ} = \frac{1}{4} \times 20 = 5 \, \text{cm}.$
Thus, the minimum length of $PQ$ to just glow the LED is:
$\ell_{PQ} = 5 \, \text{cm}.$