Question

A 100 watt, 110 volt and a 50 watt, 110 volt lamp are connected in series across 220 volt d.c. source. If the resistances two lamps are assumed to remain constant, the voltage across 100 watt lamps is

Answer 1

Hint: Derive the relation between power, current, and voltage using their mathematical expressions. Find the value for the two constants resistances. Then the total resistance of the circuit, followed by the current in the circuit. Finally, calculate the voltage across the required resistor using these values.

Formula Used:
Power, $P = I \times V$
Ohm’s law, $V = I \times R$
Series resistance, $R = {R_1} + {R_2} + {R_2} + \cdots + {R_n}$

Complete step-by-step answer:
Electrical power is defined as the rate at which work is done in an electrical circuit or the rate per unit time at which electrical energy is transferred through the electric circuit.
Mathematically, $P = I \times V \cdots \cdots \cdots \cdots \left( 1 \right)$
Ohm’s law states that the electric current in an electric circuit is directly proportional to the voltage across the circuit and inversely proportional to the resistance offered by the circuit.
Mathematically, $V = I \times R \cdots \cdots \cdots \cdots \left( 2 \right)$
Given:
The power of lamp 1, ${P_1} = 100watt$
The power of lamp 2, ${P_2} = 110watt$
The power of lamp 3, ${P_3} = 50watt$
The power of lamp 4, ${P_3} = 110watt$
The voltage of the D.C. source, ${V_ \circ } = 220V$
Lamps 1, 2, 3, and 4 are in series with each other. Additionally, it is given that the resistance across two lamps remains constant. This means that the voltage drop across the set of those two resistors will be the same.
$\eqalign{
  & \Rightarrow {V_0} = 2V \cr
  & \Rightarrow 220 = 2V \cr
  & \Rightarrow V = \dfrac{{220}}{2} \cr
  & \therefore V = 110V \cr} $
So, the resistance across lamp 1 and 2 will be given by:
$\eqalign{
  & {R_1} = \dfrac{{{V^2}}}{{{P_1}}}{\text{ }}\left[ {{\text{by equation (1) & (2)}}} \right] \cr
  & \Rightarrow {R_1} = \dfrac{{{{110}^2}}}{{100}} \cr
  & \therefore {R_1} = 121\Omega \cr} $
Similarly, the resistance across lamp 3 and 4 will be given by:
$\eqalign{
  & {R_2} = \dfrac{{{V^2}}}{{{P_1}}} \cr
  & \Rightarrow {R_2} = \dfrac{{{{110}^2}}}{{50}} \cr
  & \therefore {R_1} = 242\Omega \cr} $
The total resistance of the circuit will be the sum of these individual resistances as the connection is in series, so we have:
$\eqalign{
  & R = {R_1} + {R_2} \cr
  & \Rightarrow R = \left( {121 + 242} \right)\Omega \cr
  & \therefore R = 363\Omega \cr} $
The current is series connection remains the same, so we have:
$\eqalign{
  & {V_ \circ } = IR \cr
  & \Rightarrow I = \dfrac{{{V_ \circ }}}{R} \cr
  & \Rightarrow I = \dfrac{{220}}{{363}} \cr
  & \therefore I = 0.6A \cr} $
Now, the voltage drop across the lamp 1 will be given by:
$\eqalign{
  & {V_1} = I{R_1} \cr
  & \Rightarrow {V_1} = 0.6 \times 121 \cr
  & \therefore {V_1} = 73.3V \cr} $
Therefore, the voltage across 100 watt lamps is 73.3V.

Note: When the electrical circuit has a series connection, then the current in that region remains the same because there is only one path for the conduction of the current in that circuit. While in parallel connection the possible paths for current are multiple, but the wires are connected together which have the same voltage, so the voltage remains the same in parallel connection.