Question

If each diode in Figure has a forward bias resistance of $25\Omega$ and infinite resistance in reverse bias, what will be the values of the current $I_1,I_2,I_3$ and $I_4$ ?



Forward bias:

When the positive terminal of a diode is connected to a high voltage and the negative terminal is connected to a low voltage then the diode is forward biased. When connected in the forward bias it allows current to pass through it.

Answer 1

$\text{Draw a labelled diagram of given situation}.$


$\text{Find effective resistance of circuit}.$

Given,

Forward biases resistance $=25\Omega$

Reverse biases resistance $=\infty$

According to the figure,

The diode in branch $CD$ is in reverse biases thus having infinite resistance.

So current in that branch $CD,I_3=0$

Resistance in branch $AB\left(R_1\right)=25+125=150\Omega$

Resistance in branch $EF\left(R_2\right)=25+125=150\Omega$

As we know, $AB$ is parallel to $EF$ So effective resistance,

$\frac{1}{R^{\prime }}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{150}+\frac{1}{150}=\frac{2}{150}$

$\Rightarrow R^{\prime }=75\Omega$

$\text{Find}{I}_1.$

Total resistance $R$ of the circuit $=R^{\prime }+25=75+25=100\Omega$

$I_1=\frac{V}{R}=\frac{5}{100}=0.05A$

$\text{Apply Kirchhoff’s law and find current in branch}.$

According to the Kirchhoff's current law $\left(KCL\right)$

$I_1=I_4+I_2+I_3$

$(I_3=0)$

$I_1=I_4+I_2$

Here the resistances $R_1$ and $R_2$ is same therefore,

$I_4=I_2$

$\therefore I_1=2I_2$

$I_2=\frac{I_1}{2}=\frac{0.05}{2}=0.025A$

And $I_4=0.025A$

$\text{Final Answer}:I_2=I_4=0.025A$