Question 56: Current Electricity

Question

A cell of emf E and internal resistance r is connected to two external resistances $$\mathrm{R_1}$$ and $$\mathrm{R_2}$$ and a perfect ammeter. The current in the circuit is measured in four different situations:
(a) without any external resistance in the circuit.
(b) with resistance $$\mathrm{R_1}$$ only
(c) with $$\mathrm{R_1}$$ and $$\mathrm{R_2}$$ in series combination.
(d) with $$\mathrm{R_1}$$ and $$\mathrm{R_2}$$ in parallel combination.
The currents measured in the four cases in ascending order are

Accepted Answer

To find the current through the conductor at $$220^{\circ} \mathrm{C}$$, we need to understand the relationship between temperature and resistance, assuming that the resistance of the conductor increases linearly with temperature. This relationship is given by:
$$ R_T = R_0 (1 + \alpha T) $$
Where:
$$ R_T $$ is the resistance at temperature $$ T $$.
$$ R_0 $$ is the resistance at $$0^{\circ} \mathrm{C}$$.
$$ \alpha $$ is the temperature coefficient of resistance.
$$ T $$ is the temperature in degrees Celsius.
Since the current $$ I $$ is inversely proportional to the resistance $$ R $$ (Ohm's Law: $$ I = \frac{V}{R} $$), we'll first express the given currents $$ a $$ and $$ b $$ at the corresponding temperatures in terms of resistance:
At $$0^{\circ} \mathrm{C}$$:
$$ I_0 = a = \frac{V}{R_0} $$
At $$100^{\circ} \mathrm{C}$$:
$$ I_{100} = b = \frac{V}{R_{100}} = \frac{V}{R_0 (1 + 100\alpha)} $$
Dividing the equations for $$ b $$ by the equation for $$ a $$, we get:
$$ \frac{b}{a} = \frac{V / R_{100}}{V / R_0} = \frac{R_0}{R_0 (1 + 100\alpha)} = \frac{1}{1 + 100\alpha} $$
Solving for $$ \alpha $$, we get:
$$ 1 + 100\alpha = \frac{a}{b} $$
$$ 100\alpha = \frac{a}{b} - 1 $$
$$ \alpha = \frac{1}{100} \left(\frac{a}{b} - 1\right) $$
Now let's determine the current at $$220^{\circ} \mathrm{C}$$:
At $$220^{\circ} \mathrm{C}$$:
$$ I_{220} = \frac{V}{R_{220}} = \frac{V}{R_0 (1 + 220\alpha)} $$
Substituting the value of $$ \alpha $$ from above:
$$ I_{220} = \frac{V}{R_0 \left(1 + 220 \cdot \frac{1}{100} \left(\frac{a}{b} - 1\right)\right)} $$
$$ I_{220} = \frac{V}{R_0 \left(1 + 2.2 \left(\frac{a}{b} - 1\right)\right)} $$
$$ I_{220} = \frac{V}{R_0 \left(1 + 2.2 \frac{a}{b} - 2.2\right)} $$
$$ I_{220} = \frac{V}{R_0 \left(\frac{b + 2.2 a - 2.2 b}{b}\right)} $$
$$ I_{220} = \frac{V b}{R_0 (b + 2.2 a - 2.2 b)} $$
$$ I_{220} = \frac{V b}{R_0 (2.2 a - 1.2 b)} $$
Using $$ I_0 = \frac{V}{R_0} $$ from above:
$$ I_{220} = \frac{b I_0}{2.2 a - 1.2 b} $$
$$ I_{220} = \frac{b a}{2.2 a - 1.2 b} $$
Multiplying numerator and denominator by 5, we get:
$$ I_{220} = \frac{5 a b}{11 a - 6 b} $$
Therefore, the current through the conductor at $$220^{\circ} \mathrm{C}$$ is given by Option C:
$$ \frac{5 a b}{11 a - 6 b} $$

Answer

c < b < d < a

Answer

a < b < d < c

Answer

c < d < b < a

Answer

a < d < b < c

Current Electricity

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