Question

a uniform wire of resistance 15ohms is cut into three pieces in ratio 1:2:3 and the three pieces are connected to form a triangle. a cell of 10V potential with internal resistance 1ohm is connected across the highest of thre resistors. calculate the current through each part of the circuit.

Answer 1

Answer on Question #43573 – Physics - Electric Circuits

a uniform wire of resistance 15ohms is cut into three pieces in ratio 1:2:3 and the three pieces are connected to form a triangle. a cell of 10V potential with internal resistance 1ohm is connected across the highest of three resistors. calculate the current through each part of the circuit.

Solution:

$\mathrm{R} = 15 \mathrm{Ohm}$ - resistance of the uniform wire;

$1 \mathrm{x}: 2 \mathrm{x}: 3 \mathrm{x} - \text { ratio of the three pieces; }$

$\mathrm{U} = 10 \mathrm{~V}$ - potential of the cell;

$\mathrm{r} = 1 \mathrm{Ohm}$ - internal resistance of the cell;

Since the resistance is linearly proportional to the length of the wire $(\mathrm{R} = \rho \frac{\mathrm{L}}{\mathrm{S}}, \mathrm{L} - \mathrm{length})$ , the ratio of the three resistances of the pieces is equal to the ratio of the lengths of the pieces:

$$\begin{array}{l} R _ {1}: R _ {2}: R _ {3} = 1: 2: 3 \\ R _ {3} = 3 R _ {1} \\ R _ {2} = 2 R _ {1} \\ \end{array}$$

The sum of these resistances is equal to the resistance of the uniform ware. We have an equation:

$$\begin{array}{l} R = R _ {1} + R _ {2} + R _ {3} \\ R = R _ {1} + 2 R _ {1} + 3 R _ {1} \\ R = 6 R _ {1} \\ R _ {1} = \frac {R}{6}; R _ {2} = \frac {2 R}{6}; R _ {3} = \frac {3 R}{6} = \frac {R}{2} \\ \end{array}$$

In a series configuration, the current through all of the resistors is the same, and resultant resistance is equal to the sum of the resistances:

$$R_{12} = R_1 + R_2 = \frac{R}{6} + \frac{2R}{6} = \frac{3R}{6} = \frac{R}{2}$$

Formula for the resultant resistance for the parallel configuration:

$$\frac{1}{R_{\text{total}}} = \frac{1}{R_3} + \frac{1}{R_{12}}$$ $$R_{\text{total}} = \frac{R_3 R_{12}}{R_3 + R_{12}} = \frac{R_3 R_{12}}{R_3 + R_{12}} = \frac{\frac{R}{2} \cdot \frac{R}{2}}{\frac{R}{2} + \frac{R}{2}} = \frac{\frac{R^2}{4}}{R} = \frac{R}{4}$$

Formula for the total current through the circuit (Ohm's law):

$$I_{\text{total}} = \frac{U}{R_{\text{total}} + r} = \frac{U}{\frac{R}{4} + r} = \frac{10V}{\frac{15 \text{ Ohm}}{4} + 1 \text{ Ohm}} = 2.1 \text{ A}$$

Resistors $R_3 = \frac{R}{2}$ and $R_{12} = \frac{R}{2}$ in a parallel configuration are each subject to the same potential difference (voltage), however the currents through them add, since $R_3 = R_{12} \Rightarrow I_3 = I_2 = \frac{I_{\text{total}}}{2} = \frac{2.105 \text{ A}}{2} = 1.05 \text{ A}$ (current is halved).

In a series configuration (resistors $R_1$ and $R_2$), the current through all of the resistors is the same:

$$I_1 = I_2 = 1.05 \text{ A}$$

Answer: $I_1 = I_2 = I_3 = 1.05 \text{ A}$;

$$I_{\text{total}} = 2.1 \text{ A}$$

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