Question

A bird sitting on a single high tension wire does not get electrocuted because
A) The circuit is not complete.
B) The bird feet have an insulating covering.
C) Capacitance of the bird is too small and the line frequency is too small.
D) None of the above.

Answer 1

Hint
You will have to make use of the concept that magnitude of the current flowing through a resistor is inversely proportional to the magnitude of the resistance. In simple words, higher is the resistance, lower is the current flowing through it and vice versa. Focus on the ways in which the body of the bird can offer high resistance such that negligible current flows through it and it does not get electrocuted.

Complete step-by-step solution :
We know that Ohm’s law is stated as $V = IR$
For a constant voltage, $R \propto \dfrac{1}{I}$
If resistance is increased manifolds, the current flowing through it will be negligible. If this same principle is applied in the case of birds, we will evaluate the factors responsible for increasing bird’s resistance. The body of the bird is hollow and contains air. The capacitance of the bird is very small.
Relation between resistance offered by capacitor , that is the capacitive resistance, ${X_C}$ is given as
${X_C} = \dfrac{1}{{\omega C}}$
Where $\omega$ is the frequency and $C$ is the capacitance.
Given that the capacitance of the bird is too small and line frequency is too small, the capacitive resistance will increase to a great extent such that no current or negligible current will pass through it. So a bird sitting on a single high tension wire does not get electrocuted.
Hence, option C is correct.

Note:- Other options are incorrect as the circuit is complete. We can say so because current keeps flowing in the wire. The feet of the bird have no special covering or insulation to protect it from electrocution. Most importantly, in this case, we do not consider resistance, we focus on capacitive resistance.