Question

Three resistances, $ 1\Omega ,2\Omega ,3\Omega $ are connected in Series. What is the total resistance of the combination?
If the combination is connected to a $ 12{V} $ battery of negligible internal resistance, obtain the potential drop across each resistor.

Answer 1

Hint: The effective resistance of n Resistors of Resistances \[{R_1},{R_2},{R_3} \cdot \cdot \cdot {R_n}\]is $ {R_{eff}} = {R_1} + {R_2} + {R_3} + \cdot \cdot \cdot + {R_n} $ . Also, when the current in a circuit is the same, the Resistances and Voltage across them follow a direct proportion.

Complete step-by-step answer:
We know that when three resistors- $ {R_1} $ , $ {R_2} $ , $ {R_3} $ are connected in series, their effective resistance is given by : $ {R_{eff}} = {R_1} + {R_2} + {R_3} $
Here, Since the resistances - $ {R_1} = 1\Omega $ , $ {R_2} = 2\Omega $ and $ {R_3} = 3\Omega $ $ {R_{eff}} = 1\Omega + 2\Omega + 3\Omega = 6\Omega $ .

Now let's consider this combination of resistors being connected in series with a cell of $ 12V $ emf.

Since the cell is assumed to have zero internal resistance, the ends of the combination are directly connected to $ 12V $ emf and hence, the potential difference across the combination is $ 12V $ .
Let us now find the current passing through this circuit. Since the effective resistance of the combination is $ 6\Omega $ , The current in the circuit can be found using ohm's law as: $ I = \dfrac{V}{R} $ Substituting values of $ V $ and $ R $ gives us : $ I = \dfrac{{12}}{6} = 2A $
This $ 2A $ current is flowing through all three resistors since they are connected parallel.
Now, we are to find the potential drop across each resistor. So let the potential drop across resistors $ {R_1} $ , $ {R_2} $ , $ {R_3} $ be $ {V_1} $ , $ {V_2} $ , $ {V_3} $ respectively. Now ohm's Law tells us that the voltages $ {V_1} $ , $ {V_2} $ and $ {V_3} $ are related to the current $ I $ flowing through them and their resistances as-

$ V = I \times R $ So $ {V_1} = I \times {R_1} = 2 \times 1 = 2V $ $ {V_2} = I \times {R_2} = 2 \times 2 = 4V $ $ {V_3} = I \times {R_3} = 2 \times 3 = 6V $ So the potential difference across $ {R_1} $ is $ 2V $ , $ {R_2} $ is $ 4V $ , $ {R_3} $ is $ 6V $ .

Additional Information:
we see that summing the voltage across all resistors - $ 2V + 4V + 6V $ , we get the emf of the cell. This is true for all closed loops in a Resistive circuit and called Kirchoff's voltage rule. Also, we can see that since the current is the same in a series circuit, Ohm's law simplifies to $ V \propto R $ . So the total voltage $ 12V $ gets divided among the resistors in the direct ratio of their Resistance.

Note: For competitive examinations, calculations have to be simplified and hence, we can use the Direct proportionality between $ V $ and $ R $ to find the Voltages.
Since $ {V_1}:{V_2}:{V_3}::1:2:3 $
 $ {V_1} = \dfrac{1}{{1 + 2 + 3}}{V_{tot}} $
 $ {V_1} = \dfrac{1}{6}12V = 2V $
Similarly,
 $ {V_2} = \dfrac{2}{6}12V = 4V $
 $ {V_3} = \dfrac{3}{6}12V = 6V $