Question

The drawing shows three different resistors in two different circuits. the battery has a voltage of v = 23 v, and the resistors have resistances of r1 = 50.0 ?, r2 = 25.0 ? and r3 = 10.0 ?. determine the current through and the voltage across each resistor.

Answer 1

Missing picture in the text. I've found it here:
https://www.flickr.com/photos/[email protected]/6791933847/in/photostream

Solution:

Circuit 1)
The three resistors are in series. So the equivalent resistance of the circuit is
$R_{eq}=R_1+R_2+R_3 =50.0\Omega+25.0 \Omega+10.0 \Omega=85.0 \Omega$
So the current flowing through each resistor is the same, and it is given by Ohm's law:
$I= \frac{V}{R_{eq}}= \frac{23 V}{85.0 \Omega}=0.27 A$

And the voltage drop across each resistor is given by Ohm's law as well:
$V_1 = I R_1 = (0.27 A)(50.0 \Omega)=13.5 V$
$V_2 = I R_2 = (0.27 A)(25.0 \Omega)=6.8 V$
$V_3 = I R_3 = (0.27 A)(10.0 \Omega)=2.7 V$


Circuit 2)
The three resistors are in parallel, so their equivalent resistance is given by
$\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{50\Omega}+ \frac{1}{25 \Omega}+ \frac{1}{10 \Omega}= \frac{7}{50 \Omega}$
$R_{eq}= \frac{50}{7} \Omega=7.1 \Omega$

The voltage drop across each resistor is the same, and it is equal to the voltage of the battery:
$V_1 = V_2 = V_3 = 23 V$

while the current across each resistor is given by Ohm's law:
$I_1 = \frac{V}{R_1}= \frac{23 V}{50 \Omega}=0.46 A$
$I_2 = \frac{V}{R_2}= \frac{23 V}{25.0 \Omega}=0.92 A$
$I_3 = \frac{V}{R_3}= \frac{23 V}{10 \Omega}=2.3 A$