Question

Find the value of i₁/i₂ in the following figure if (a) R = 0.1 Ω (b) R = 1 Ω and (c) R = 10 Ω. Note from your answers that in order to get more current from a combination of two batteries, they should be joined in parallel if the external resistance is small and in series if the external resistance is large, compared to the internal resistance.

Answer 1

(a) For R = 0.1 Ω

Applying KVL in the given circuit, we get:-

\[0 . 1 i_1  + 1 i_1  - 6 + 1 i_1  - 6 = 0\]

\[ \Rightarrow 0 . 1 i_1  + 1 i_1  + 1 i_1  = 12\]

\[ \Rightarrow  i_1  = \frac{12}{\left( 2 . 1 \right)} = 5 . 71  A\]

Now, consider the given circuit.

Applying KVL in the loop ABCDA, we get:-

\[0 . 1 i_2  + 1i - 6 = 0\]

\[ \Rightarrow 0 . 1 i_2  + i = 6\]

\[ \Rightarrow i = 6 - 0 . 1 i_2\]

Applying KVL in ADEFA, we get:-

\[i - 6 + 6 - \left( i_2 - i \right)1 = 0\]

\[ \Rightarrow i -  i_2  + i = 0\]

\[ \Rightarrow 2i -  i_2  = 0\]

\[ \Rightarrow 2\left[ 6 - 0 . 1  i_2 \right] -  i_2  = 0\]

\[ \Rightarrow    i_2  = 10  A\]

\[\therefore   \frac{i_1}{i_2} = 0 . 571\]

 

(b) For R = 1 Ω

Applying KVL in the circuit given in figure 1, we get:-

\[1 i_1  + 1 .  i_1  - 6 +  i_1  - 6 = 0\]

\[ \Rightarrow 3 i_1  = 12\]

\[ \Rightarrow  i_1  = 4\]

Now, for figure 2:-

Applying KVL in ABCDA, we get:-

\[i_2  + i - 6 = 0\]

\[ \Rightarrow    i_2  + i = 6\]

Applying KVL in ADEFA, we get:-

\[i - 6 + 6 - \left( i_2 - i \right)1 = 0\]

\[ \Rightarrow i -  i_2  + i = 0\]

\[ \Rightarrow 2i -  i_2  = 0\]

\[ \Rightarrow 2\left[ 6 - i_2 \right] -  i_2  = 0\]

\[ \Rightarrow   12 - 3 i_2  = 0\]

\[ \Rightarrow  i_2  = 4  A\]

\[ \therefore \frac{i_1}{i_2} = 1\]

(c) For R = 10 Ω

Applying KVL in the circuit given in figure 1, we get:-

\[10 i_1  + 1 i_1  - 6 + 1 i_1  - 6 = 0\]

\[\Rightarrow 12 i_1  = 12\]

\[ \Rightarrow  i_1  = 1\]

Now, for figure 2:-

Applying KVL in ABCDA, we get:-

\[10 i_2  + i - 6 = 0\]

\[ \Rightarrow i = 6 - 10 i_2\]

Applying KVL in ADEFA, we get:-

\[i - 6 + 6 - \left( i_2 - i \right)1 = 0\]

\[ \Rightarrow i -  i_2  + i = 0\]

\[ \Rightarrow 2i -  i_2  = 0\]

\[ \Rightarrow 2\left[ 6 - 10 i_2 \right] -  i_2  = 0\]

\[ \Rightarrow   12 - 21 i_2  = 0\]

\[ \Rightarrow  i_2  = 0 . 57  A\]

\[ \therefore \frac{i_1}{i_2} = 1 . 75\]