A generator has an emf of 440 volt and internal resistance 400 ohm. Its terminals are connected to a load of 4000 ohm. The voltage across the load is:
(A) 220 volt
(B) 440 volt
(C) 200 volt
(D) 400 volt

Hint: When the current flows through the battery it flows due to flow of ions inside the chemicals of the battery rather than drift of the free electrons in the conducting wires. Certainly the mobile for the flow of ions is less compared to mobility of free electrons in conductors so the conductivity of chemicals is less compared to connecting wires hence the battery chemicals always offer some resistance to flow of current as it is connected in series. This resistance due to the chemical composition of the battery is called ‘Internal resistance of the battery’.

Complete step by step solution:
Given:
Internal Resistance, r = 400 Ω
Resistance of the load, R = 4000 Ω
Voltage of the generator, V = 440 V
Total resistance \[{R_T} = (4000 + 400)\Omega = 4400\Omega \]
Current flowing, \[I = \dfrac{V}{{{R_T}}}\]
\[ \Rightarrow I = (\dfrac{{440}}{{4400}})A\]
\[\therefore I = 0.1A\]
Now,
Load Voltage, \[{V_L} = I{R_T}\]
\[ \Rightarrow {V_L} = (0.1 \times 4000)V\]
\[\therefore {V_L} = 400V\]

Hence Option (D) is the correct answer.

Note: In grouping of cells or electronic devices resistance of the circuit will be the combination of the resistances offered by all the devices and the internal resistances of the cells whether it is given or not. The combination resistances will be either in series or in parallel it depends upon the addition of resistors.