Question

Find the circuit in the three resistors shown in the figure (32-E15).
figure 32-E15

Answer 1

Answer:

From the figure, we can see that current will flow only in loop 1 because current follows the least resistive path. All the current will pass through the wire connected in parallel to the 4 Ω resistor in loop 2.

∴ There will be no current through the 4 Ω resistor in loop 2.

For loop1, applying KVL, we get:

6i+4i−4+2=0

⇒10i = 2

⇒i = 0.2 A

∴ The current through the 4 Ω and 6 Ω resistors, i = 0.2 A

Answer 2

Explanation:

When loop 1 is applied in KVL, we obtain:

2 + i1 - i2 $\times$ 1 – 2 = 0

⇒i1 = i2

In loop 2, when KVL is applied, we obtain:

2 + i2 - i3 $\times$ 1 – 2 - i1 - i2 $\times$ 1 = 0

⇒i2 - i3 - i1 + i2 = 0 i1 = i2

⇒i2 - i3 - i2 + i2 = 0

⇒i2 = i3

In loop 3, when KVL is applied, we obtain:

2 + i3 – 2 - i2 - i3 = 0

⇒i3 = 0 i1 = i2 = i3

Therefore, i1 = i2 = i3 = 0.