Question

If break down voltage of zener diode is 6 V then value of I2 will be :- ​

Answer 1

The value of I₂ will be 0.018 A

Given,

Value of the voltage drawn from the main battery is (V) = 18 V

The value of resistance attached in series with the circuit (R$_i$) = 500 Ω

Current passing through R$_i$ is = I

The value of load resistance (R$_L$) = 1 kΩ

                                                       = 10³ Ω

The amount of current passing through R$_L$ is = I₁

The breakdown voltage of zener diode (V$_Z$) = 6 V

The current passing through the zener diode is = I₂

• The zener diode acts as a voltage stabilizer.
• So, the amount of current which is passing through it is = Total current passing through the circuit - current passing through the  parallel circuit of zener diode.

Now,

From Ohm's Law, we know,

 Voltage = Current × Resistance

   ∴ I = 18/500 A

          = 36/ 1000 A

And,

   ∴ I₁ = 18/10³ A

  ∵ I₂ = I - I₁,

∴ I₂ = 18/1000 A

      = 0.018 A

So, the value of I₂ will be 0.018 A.