When a high resistance voltmeter is connected directly across an electric bulb,its reading is $2 \ V$. An electric cell is sending a current of $0.4 \ A$ (measured by an ammeter) in the electric circuit.
$(a)$ Draw the circuit.
$(b)$ Find the resistance of the electric bulb.
$(c)$ State the law that is applied for making these calculations. If a graph is plotted between $V$ and $I$,show the nature of the graph obtained.

(B) The circuit diagram consists of an electric bulb,a voltmeter connected in parallel to the bulb,an ammeter connected in series,a cell,and a key,as shown in the provided image.
$(b)$ Given:
Current $(I) = 0.4 \ A$
Potential difference $(V) = 2 \ V$
According to Ohm's law,$V = IR$
Therefore,$R = \frac{V}{I} = \frac{2 \ V}{0.4 \ A} = 5 \ \Omega$
So,the resistance of the electric bulb is $5 \ \Omega$.
$(c)$ The law applied is Ohm's law. The $V-I$ graph for an ohmic conductor is a straight line passing through the origin,as shown in the provided image.