Question

Four resistors are arranged in a circuit, as shown. There is aconstant voltage V across the voltage source. The current I is 0.25 ampere,(a) Find the equivalent resistance of the four resistors.(b) Find the voltage V across the voltage source.(c) Find the voltage across resistor R2.(d) Find the power dissipated by resistor R2.

Answer 1

We are given 2 resistors in parallel and 3 resistors in series. The equivalent resistance for parallel resistors is determined using the following formula:

`(1)/(R_(23))=(1)/(R_2)+(1)/(R_3)`

Solving for the equivalent resistance:

`R_(23)=(R_3R_2)/(R_3+R_2)`

Replacing the values we get:

`R_(23)=((40)(60))/(40+60)`

Solving the operations:

`R_(23)=24`

Now we use this resistance in series with resistances 1 and 4 to determine the equivalent resistance of the circuit using the following formula:

`R_(1234)=R_1+R_(23)+R_4`

Replacing the values:

`R_(1234)=16+24+12`

Solving the operations:

`R_(1234)=52`

Therefore, the equivalent resistance of the 4 resistors is 52 Ohm.

For part B we are asked to determine the voltage V across the circuit. To do that we will use Ohm's law, which is:

`V=IR`

Where "I" is the current and "R" is the equivalent resistance. Replacing the values:

`\begin{gathered} V=(0.25A)(52\text{Ohm)} \\ V=13V \end{gathered}`

Therefore, the voltage across the circuit is 13 Volts.

For part C. We will use Kirchhoff law to determine the current in the second loop of the circuit. This says that the sum of the voltages in a loop must be equal to zero, this means for the first loop:

`V-R_1I-I_2R_2-I_{}R_4=0`

Now, since the drop in voltage is equivalent to:

`V_2=I_2R_2`

We can solve for this value in Kirschhoff's law:

`I_2R_2=V-IR_4-IR_1`

Replacing the values:

`V_2=13-(0.25)(12)-(0.25)(16)`

Solving the operations:

`V_2=6`

Therefore, the voltage drop in resistance 2 has a magnitude of 6 volts.

For part D, we will use the following formula:

`P=VI`

Where "P" is the power dissipated, "V" the voltage drop, and "I" the current, since we don't know the current but the resistance we can rewrite the formula as:

`P=(V^2)/(R)`

Replacing the values:

`P_2=((6)^2)/(60)=0.6`

Therefore, the power dissipated in resistance 2 is 0.6 J.