Question

In the circuit diagram, calculate: (a) effective resistance (b) current drawn from cell [Ans.(a) 32. (b) 0.5 A]​

Answer 1

Effective resistance is 3 Ω

Current in the circuit 0.5 A

Step-by-step explanation:

To find effective resistance;

Two resistance of 2 Ω are in parallel, solving it to get a single resistance:

$\frac{1}{R} = \frac{1}{R_1} +\frac{1}{R_2}$

$\frac{1}{R} = \frac{1}{2} +\frac{1}{2}$

$\frac{1}{R} = \frac{1+1}{2} = \frac{2}{2} = 1$

R = 1 Ω

Now, three 1ohms resistance are in series

$R_{eff} = R_3 + R_4 + R$

[tex]R_{eff} = 1+1+1 = 3\\ R_{eff} = 3 ohms[/tex]

To calculate current:

[tex]V = IR_{eff}\\ [/tex]

Given: V = 1.5V

[tex]1.5= I * 3\\ I = \frac{1.5}{3} = \frac{1}{2} = 0.5\\ I = 0.5 A [/tex]

Answer 2

Answer: effective resistance 3 and current drawn 0.0.5A

Step-by-step explanation:2ohm are in parallel so effective resistance 1 then all three are in series so equivalent resistance 1+1+1=3..then ohms law V=IR so current drawn I= V/R =0.0.5