Find the current in each branch of the given network if the total current is 2.55 A.

Concept:

Different connections of resistance.

1.Series Connection.

R_eq = R₁ + R₂ + R₃

If N resistance is connected in series then, 

R_eq = R₁ + R₂ + R₃

If N resistance is connected in series then, 

R­_eq = R₁ + R₂ + R₃ + ------ + R_N-1 + R_N = $\sum _{i=1}^N{R}_i$

2 Parallel Connection

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

• If N resistance are connected in parallel then,

 $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+----+\frac{1}{R_{N-1}}+\frac{1}{R_N}=\sum _{i=1}^N\frac{1}{R_i}$

Current Division Rule:

The current in any of the parallel branches is equal to the ratio of opposite branch resistance to the total resistance, multiplied by the total current.

$I_1=\frac{R_2}{R_1+R_2}I\mathrm{\&}{I}_2=\frac{R_1}{R_1+R_2}I$

Calculation:

Here, V = 10 V

R_a = 2 Ω

R_b = 8 Ω  

R_c = 5 Ω

R_x = R_a + R_b = 2 + 8 = 10 Ω

R_y = R_c + R = (5 + R) Ω 

R_eq = R_x || R_y =$\frac{\mathrm{R}\mathrm{x}\mathrm{R}\mathrm{y}}{\mathrm{R}\mathrm{x}+\mathrm{R}\mathrm{y}}=\frac{10\left(5+R\right)}{10+5+R}=\frac{10\left(5+R\right)}{15+R}$

Since, V = IR_eq

_{$10=2.55\times \frac{10\left(5+R\right)}{15+R}$}

$\frac{15+R}{5+R}=2.55$

15 + R = 12.75 + 2.55R

1.55R = 2.25

R = 1.45 Ω

So R_y = 5 + R = 5 + 1.45 = 6.45 Ω

Now, through current division Rule 

$I_1=\frac{R_y}{R_x+R_y}\times I$

 =$\frac{6.45}{10+6.45}\times 2.55$

= $\frac{6.45\times 2.55}{16.45}=1A$

$I_2=\frac{R_x}{R_x+R_y}I$

= $\frac{10}{10+6.45}2.55$

= $\frac{10\times 2.55}{16.45}=1.55A$

Important Points

Voltage Division Rule:

Voltage across a resistor in a series circuit is equal to the value of that resistor times the total voltage across the series elements divided by the total resistance of the series elements.

$V_{R1}=\frac{R_1}{R_1+R_2}{V}_s\mathrm{\&}{V}_{R2}=\frac{R_2}{R_1+R_2}{V}_s$

Shortcut Trick:

Resistance R_x = 10 Ω

The voltage across R_x is 10 V

So current $I_1=\frac{V}{R_x}=\frac{10}{10}=1A$

From the 4 options given below, only option (2) contains I₁ = 1 Amp.

So option (2) is the correct option.