Question

Three resistors, 4ohms, 6ohms, 8ohms are connected in a parallel circuit. What is their total resistance

Answer 1

The total resistance of the three resistors connected in series is 1.85Ω.

Explanation:

Whenever a passive device is connected in a circuit, if they are connected  parallel to each other, then the voltage drop in each of the passive device will be constant. So in this case, the passive device is the resistor. So three resistors are connected parallel to each other, this means the voltage drop across these three resistors will be constant but the current flow in the three resistors will be different.

So, the net current flow in the circuit will be

$I = I_{1} + I_{2} + I_{3}$

From, Ohm's law, we know that $I = \frac{V}{R}$

So, since the voltage drop across the three resistors are same as V, then resistor be denoted as R₁ = 4 Ω R₂ = 6Ω  and R₃= 8Ω.

So, the formula becomes,

$\frac{V}{R_{p} } =\frac{V}{R_{1} } + \frac{V}{R_{2} } + \frac{V}{R_{3} }$

$\frac{1}{R_{p} } =\frac{1}{R_{1} } + \frac{1}{R_{2} } + \frac{1}{R_{3} }\\\\\frac{1}{R_{p} } =\frac{1}{4} + \frac{1}{6 } + \frac{1}{8 }\\\\\frac{1}{R_{p} } =\frac{6+4+3}{24 }$

$R_{p} = \frac{24}{6+4+3} =\frac{24}{13} =1.85$

Thus, the total resistance of the three resistors connected in series is 1.85Ω.