Question

1. What is the equivalent resistance in this circuit? 2. What is the current in this circuit?

Answer 1

1. The equivalent resistance in a series circuit, like the one shown, is given by the general equation

$R_{S} = R_1 + R_2 + R_3 + . . . + R_{N-1} + R_N = \sum_{i=1}^N R_i$

where N is the number of resistors in series. Here, there are four resistors in series. Using our equation with N = 4, we obtain

$R_{S} &= R_1 + R_2 + R_3 + R_4 \\ R_{S} = 2.0 \, \Omega + 3.0 \, \Omega + 4.0 \, \Omega + 6.0 \, \Omega \\ R_{S} = 15.0 \, \Omega.$

2. The current in this series circuit is equal to the voltage divided by the equivalent resistance (a relationship you can derive from Ohm's law):

$I = \frac{V}{R_{S}} = \frac{45 \, V}{15.0 \, \Omega} = 3.0 \, A. \nonumber$