homework_09_electric_current
A circuit is constructed with five resistors and a battery as shown. The battery voltage is [V = 12.0 V]. The values for the resistors are: [R_1 = 75.0 \Omega], [R_2 = 118.0 \Omega], [R_3 = 147.0 \Omega] and [R_4 = 61.0 \Omega].
The value for [R_X] is unknown, but it is known that [I_4], the current that flows through resistor [R_4], is zero.
Let
- [V_b = 12.0 V]
- [R_1 = 75.0 \Omega]
- [R_2 = 118.0 \Omega]
- [R_3 = 147.0 \Omega]
- [R_4 = 61.0 \Omega].
- [I_4 = 0]
What is [I_1], the magnitude of the current that flows through the resistor [R_1]?
-
[I_1 = \frac{ V_b}{ R_1 + R_3}]
-
[\frac{ v}{ R_1}]
- It looks like you have calculated the current assuming that the voltage across [R_1] is the battery voltage, which it is not. The current does not return directly to the battery after leaving [R_1]. Where does it go before returning to the battery?
What is [V_2], the magnitude of the voltage across the resistor [R_2]?
-
[I_2]
- [v_{drop} = I_1 R_1]
-
[0]
- The voltage across [R_2] is not equal to zero since the current through [R_2] is not zero. Look at the circuit closely to determine how the voltage acrosss and current through [R_1] are related to the same quantities for [R_2].
-
[12V]
- The voltage across [R_2] is not equal to the battery voltage. Think about the voltage across [R_4]. What does that tell you about [V_1] and [V_2]?