homework_09_electric_current

Homework: Electric Current

Circuit 1 with Resistor and a Battery

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]

1)

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}]

  • Correct

• [\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?

2)

What is [V_2], the magnitude of the voltage across the resistor [R_2]?

• [V_2 = V_b - I_1 R_3]

  • Since [I_4 = 0] we are able to say [V_1 = V_2].

• [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]?

3)

What is [I_2], the magnitude of the current that flows through the resistor [R_2]?

• [I_2 = \frac{ V_2}{ R_2}]

4)

What is [R_X], the value of the unknown resistor [R_X]?

• [R_X = \frac{ I_1 R_3}{ I_1}]

5)

What is [V_1], the magnitude of the voltage across the resistor [R_1]?

• [V_1 = V_2]
  • Since [I_4 = 0] we are able to say [V_1 = V_2].

6)

If the value of the resistor [R_2] were doubled, how would the value of the resistor [R_3] have to change in order to keep the current through [R_4] equal to zero?

• [R_3] would need to be decreased
  • [I_2 = \frac{ V_2}{ R_2}] would be decreased by half if [R_2] were doubled, hence [R_3] needs to be decreased.

Circuit 2 with Resistors and a Battery

A circuit is constructed with five resistors and a battery as shown. The values for the resistors are: [R_1 = R_5 = 63.0 \Omega], [R_2 = 81.0 \Omega], [R_3 = 64.0 \Omega], and [R_4 = 130.0 \Omega]. The battery voltage is [V = 12.0 V].

Let

• [V_0 = V = 12.0 V]
• [R_1 = R_5 = 63.0 \Omega]
• [R_2 = 81.0 \Omega]
• [R_3 = 64.0 \Omega]
• [R_4 = 130.0 \Omega]

1)

What is [R_{ab}], the equivalent resistance between points [a] and [b]?

• [R_{ab} = \left(\frac{ 1}{ R_2 + R_3} + \frac{ 1}{ R_4}\right)^{-1}]

  • Correct

• [R_2 + R_3]

  • It looks like you have calculated the equivalent resistance of [R_2] and [R_3] in series. These resistors are not the only resistors connected between [a] and [b]. Try redrawing the circuit after finding [R_{23}] to see what else you must take into account.

2)

What is [R_{ac}], the equivalent resistance between points [a] and [c]?

• [R_{ac} = R_1 + R_{ab} = R_1 + \left(\frac{ 1}{ R_2 + R_3} + \frac{ 1}{ R_4}\right)^{-1}]

3)

What is [I_5], the current that flows through resistor [R_5]?

• [I_5 = I_{equiv} - I_{ac}]
  • [I_{equiv} = \frac{ V_0}{ R_{equiv}}]
    • R_{equiv} = R_1 + \left(\frac{ 1}{ R_2 + R_3} + \frac{ 1}{ R_4}\right)^{-1} + R_5}
  • [I_{ac} = I_1 - I_{bc} ]
  • [V_b = V_0 - V_1]
    • [V_1 = R_1 I_1 = R_1 \frac{ V_0}{ R_equiv}]
  • [V_a = V_0 - V_1 - V_4]
    • [V_4 = R_4 I_4 = R_4 \frac{ V_b}{ R_4 + R_5 }]

5)

What is [I_1], the current that flows through the resistor [R_1]?

• [I_1 = \frac{ V_0}{ R_equiv}]