# Circuits

Multiple choice:

1. An ideal source of emf is connected in seriew with a resistor in closed loop, and a current or 0.e50 A circulates. An additional 6.0-Ohm resistor is added in series, and the current drops to 0.30 A. The original resistance was: (a) 9.0 Ohm (b) 6.0 Ohm (c) 15 Ohm (d) 3.0 Ohm (e) none of these. Explain your choice.

2. A 24-W, 12-V ohmic lamp is placed in a circuit with 6 V across it, at which tiem it draws a current of: (a) 2 A (b)1 A (c) 0 (d) 1/2 A (e) none of these. Explain your choice.

3. A battery of emf 12 V and internal resistance of 0.2 Ohm is placed across a variable resistor. The resistor is adjusted until it dissipates a maximum amount of power, at which point the current through it is: (a) 12 A (b) 0.4 A (c) 30 A (d) 60 Ohm (e) none of these. Explain your choice.

4. Two resistors of 5 Ohm and 20 Ohm are connected in parallel across and ideal source of 20 V. The current supplied by the source is: (a) 4 A (b)5 A (c) 20 A (d) 1 A (e) none of these. Explain your choice.

Problem

1. A battery with a termnal voltage of 11.95 V when delivering 0.50 A, has an internal resistance of 0.10 Ohm. What is its emf?

2. Determine teh termnal voltage of a battery having an emfof 12.0 V and an internal resistance of 0.20 Ohm, knowing that when placed across its terminals, a 40.0-Ohm resistor dissapates 10.0 W.

3. Two 30.0-Ohm resistors are attached in series and then placed across the termnals of a 12.0-V battery having a negligible internal resistance. How much current flows through the circuit?

4. Three resistors with values of 2.0 Ohm, 3.0 Ohm, and 6.0 Ohm are connected in parallel. What is the equivalent resistance?

5. A resistor is placed in series with an uncharged 2.0-uF capacitor and a 12.0-V battery is put across the two. If the current that immediately flows around the circuit is measured to be 12 uA, determine the resistance. What is the time constant of the circuit?

#### Solution Preview

<br>For all the problems below use the basic laws of electricty:

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<br>Ohm's Law: Voltage = Current * Resistance, V = I * R

<br>Resistances in Series: Effective Resistance Reff = Sum of all individual resistances

<br> = R1 + R2 + R3+ .....

<br>Resistances in Parallel: Effective Resistance Reff is given by:

<br> 1/Reff = 1/R1 + 1/R2+1/R3+.......

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<br>Power (in Watts) = Voltage (in Volts) * Current (in Amps)

<br> P = V * I

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<br>now apply ohm's law, P = I (square) R = I * I * R and

<br> P = V (square) / R = v* V / R

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<br>Multiple choice:

<br>1. An ideal source of emf is connected in seriew with a resistor in closed loop, and a current

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<br>or 0.e50 A circulates. An additional 6.0-Ohm resistor is added in series, and the current drops

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<br>to 0.30 A. The original resistance was: (a) 9.0 Ohm (b) 6.0 Ohm (c) 15 Ohm (d) 3.0 Ohm (e) none

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<br>of these. Explain your choice.

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<br>ANS: I am assuming the current is 0.5 A (You had indicated it as 0.e50 A).

<br> Answer is (A) 9.0 Ohms

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<br>Apply Ohm's Law: Voltage = Current * Resistance, V = I * R

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<br>Call the emf as V Volts and the original Resistance as R Ohms.

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<br>Apply the Ohm's law to Original circuit: V = 0.5 * R Eq. (1)

<br>Apply Ohm's law to the new circuit: V = 0.3 * ( R+6) Eq. (2)

<br>Divide (1) by (2) and solve for R and you get R = 9 Ohms

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<br>2. A 24-W, 12-V ohmic lamp is placed in a circuit with 6 V across it, at which tiem it draws ...