Science and Mathematics Education Research Group FACULTY OF EDUCATION Department of Curriculum and Pedagogy Physics Circuits Science and Mathematics Education Research Group Categories: Secondary – Physics – Circuits Tags: voltage, current, resistance, Ohm’s law, Kirchhoff’s laws Excerpt: A guided explanation of potential differences and how they can be used in solving a series of increasingly convoluted electrical circuits. Supported by UBC Teaching and Learning Enhancement Fund 2012-2013

Question Title Question Title Multiple Battery Circuit R For additional resources on circuits and circuit simulations, check: PhET simulations: Circuit Construction Kit (DC Only) - http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc and Circuit Construction Kit (DC and AC): http://phet.colorado.edu/en/simulation/circuit-construction-kit-ac To see simulations of circuit: http://falstad.com/circuit/e-index.html . Make sure you have java installed or the applet will not run. R

Question Title Question Title Single Battery Circuit I What is the current through the resistor labelled R? Each resistor has a resistance of 1 Ω and each battery is a 1 V battery. 0.5 A 1 A 2 A 4 A None of the above e = 1 V R Circuit built on falstad.com/circuit

Comments Comments Solution Answer: B Justification: This is a simple one battery, one resistor (one loop) circuit. According to Ohm’s Law: , the current across the resistor is simply the voltage provided by the battery divided by the resistance: All other answers are incorrect application of the Ohm’s Law. Additional Resources: To see a simulation of this circuit, go to goo.gl/W4D3c. Make sure you have java installed or the applet will not run or you can alternatively use PhET circuit simulations.

Question Title Question Title Multiple Battery Circuit II What is the current through the resistor labelled R1? Each resistor is 1 Ω and each battery is 1 V 0.5 A 1 A 2 A 4 A None of the above R1 = 1 W A B e = 1 V R2 = 1 W Circuit built on falstad.com/circuit

Comments Comments Solution Answer: B Justification: Since VB-VA is equal to the battery voltage, or 1 V, the voltage across resistor R1 is 1 V. Therefore, applying the same considerations as in the earlier question (Ohm’s law), 1 A of current must flow across resistor R1. The rest of the circuit is superfluous as we only need to consider the potential difference across R1. Additional Resources: To see a simulation of this circuit, go to goo.gl/IZemE. All other answers apply Ohm’s law incorrectly. Key concept: If you know the voltage across a resistor and its resistance, you can use Ohm’s Law to find the current flowing through it. A B R1

Question Title Question Title Multiple Battery Circuit III What is the current through the resistor labelled R1? Each resistor is 1 Ω and each battery is 1 V 4 A 3.45 A 2 A 1 A None of the above R1 = 1 W Circuit built on falstad.com/circuit

Comments Comments Solution Answer: C Justification: Similar to the last question, the top part of this circuit is completely irrelevant. As the voltage at the terminals of a battery must differ by the value of the battery, the voltage between the ends of the resistor must be equal to 2 V as the resistor is connected to two batteries in a loop. This gives us our answer of 2 A for the current. Key concept: If you know the voltage across a resistor and its resistance, you can use Ohm’s Law to find the current flowing through it. In a simple case, in order to find this voltage try to identify a closed loop that includes this element, batteries and no other resistors. Remember to notice how the batteries are connected (directionality) – “+ and –” . Additional Resources: To see a simulation of this circuit, go to goo.gl/xY2tE. A B R1

Question Title Question Title Multiple Battery Circuit IV What is the current through the resistor labelled R1? Each resistor is 1 Ω and each battery is 1 V 3 A 7 A 3.9 A 1 A None of the above R1 = 1 W Circuit built on falstad.com/circuit

Comments Comments Solution Answer: A Justification: Look at the outside loop of the circuit, which is three batteries and a resistor. Each battery brings the voltage up by 1 V. Since there are three batteries, there is 3 V across the resistor, which means that 3 A flows through the resistor. Key concept: If you know the voltage across a resistor and its resistance, you can use Ohm’s Law to find the current flowing through it. In a simple case, in order to find this voltage try to identify a closed loop that includes this element, batteries and no other resistors. Remember to notice how the batteries are connected (directionality) – “+ and –” . Additional Resources: To see a simulation of this circuit, go to goo.gl/WaKRg. A B R1

Question Title Question Title Multiple Battery Circuit V What is the current through the resistor labelled R1? Each resistor is 1 Ω and each battery is 1 V 3 A 7 A 3.9 A 1 A None of the above R1 = 1 W Circuit built on falstad.com/circuit

Comments Comments Solution Answer: A Justification: There is a loop consisting only of three batteries and the resistor R1. Since the potential difference across ends of the resistor must be equal to the potential difference across the three batteries in the loop, 3 A of current flows through the resistor due to the 3 V of potential difference. Additional Resources: To see a simulation of this circuit, go to goo.gl/l00S2. Key concept: While the question looks more complicated than the questions we encountered earlier, the concept behind it is the same and it is NOT more difficult than the previous ones! R1 A B

Question Title Question Title Multiple Battery Circuit VI What is the current through the resistor labelled R1? Each resistor is 1 Ω and each battery is 1 V. 3 A 7 A 3.9 A 1 A None of the above Notice the direction of the battery is reversed compared to the previous question R1 = 1 W Circuit built on falstad.com/circuit

Comments Comments Solution Answer: A Justification: There is a loop consisting only of three batteries and the resistor R1. Once again, the potential difference across ends of the resistor must be equal to the potential difference across the three batteries, which is 1 V in this case: 1 V – 1 V +1 V = 1 V (notice the directionality of the middle battery). Therefore, the current of 1 A flows through the resistor . Key concept: Notice, while adding up the voltages across the batteries, you have to pay attention at the directions of the batteries relatively to each other: two identical batteries connected in series can either enhance each other: 1 V + 1 V = 2 V, or cancel each other 1 V – 1 V = 0 V (as in this example). Additional Resources: To see a simulation of this circuit, go to goo.gl/l00S2. R1 A B

Question Title Question Title Multiple Battery Circuit VII What is the current through the resistor labelled R? Each resistor is 1 Ω and each battery is 1 V 15.42 A 11 A 0 A 3.97 A None of the above R Circuit built on falstad.com/circuit

Comments Comments Solution Answer: B Justification: Consider the highlighted route below, which consists of 11 batteries and resistor R. As you can see, since each battery is connected to another by a wire, the highlighted network has a total potential difference of 11 V at the ends of resistor R. 11 V/1 Ω is 11 A, which is our answer. http://falstad.com/circuit/#%24+17+5.0E-6+10.20027730826997+25+5.0+50%0Ar+48+128+48+16+0+1.0%0Aw+272+32+336+32+0%0Ar+336+32+448+16+0+1.0%0Ar+48+128+128+224+0+1.0%0Aw+48+16+128+224+0%0Ar+128+224+32+240+0+1.0%0Av+48+128+32+240+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+32+240+32+288+0+1.0%0Ar+128+224+128+272+0+1.0%0Ar+128+272+32+288+0+1.0%0Ar+128+272+144+336+0+1.0%0Ar+144+336+176+384+0+1.0%0Ar+176+384+112+400+0+1.0%0Av+32+288+112+400+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+112+400+224+448+0+1.0%0Av+176+384+224+448+0+0+40.0+1.0+0.0+0.0+0.5%0Av+144+336+224+304+0+0+40.0+1.0+0.0+0.0+0.5%0Aw+176+384+256+368+0%0Ar+256+368+224+448+0+1.0%0Ar+256+368+224+304+0+1.0%0Ar+224+304+256+240+0+1.0%0Av+128+272+256+240+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+272+32+288+112+0+1.0%0Ar+288+112+336+32+0+1.0%0Av+272+32+224+80+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+224+80+160+112+0+1.0%0Ar+160+112+128+224+0+1.0%0Ar+272+32+48+16+0+1.0%0Av+128+224+288+112+0+0+40.0+1.0+0.0+0.0+0.5%0Av+160+112+48+16+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+288+112+256+240+0+1.0%0Ar+256+240+320+224+0+1.0%0Aw+288+112+320+224+0%0Ar+288+112+336+128+0+1.0%0Ar+336+128+320+224+0+1.0%0Ar+400+176+384+96+0+1.0%0Av+336+128+384+96+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+320+224+400+176+0+1.0%0Av+560+48+448+16+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+496+176+560+48+0+1.0%0Ar+496+176+448+16+0+1.0%0Aw+400+176+496+176+0%0Av+496+176+336+32+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+560+48+576+128+0+1.0%0Ar+576+128+576+192+0+1.0%0Ar+576+192+624+224+0+1.0%0Ar+496+176+464+256+0+1.0%0Ar+464+256+528+256+0+1.0%0Ar+528+256+560+256+0+1.0%0Av+560+256+576+192+0+0+40.0+1.0+0.0+0.0+0.5%0Av+320+224+256+368+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+256+368+320+352+0+1.0%0Ar+320+352+336+384+0+1.0%0Av+320+352+384+320+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+384+320+416+352+0+1.0%0Ar+336+384+400+400+0+1.0%0Ar+400+400+224+448+0+1.0%0Aw+416+352+400+400+0%0Aw+336+384+224+448+0%0Av+336+384+416+352+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+384+320+320+224+0+1.0%0Ar+320+224+464+256+0+1.0%0Ar+464+256+384+320+0+1.0%0Av+400+400+528+256+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+400+400+544+368+0+1.0%0Aw+528+256+544+368+0%0Ar+400+400+480+416+0+1.0%0Ar+544+368+608+416+0+1.0%0Av+480+416+608+416+0+0+40.0+1.0+0.0+0.0+0.5%0Av+544+368+624+224+0+0+40.0+1.0+0.0+0.0+0.5%0Ar+624+224+608+416+0+1.0%0Av+464+256+400+176+0+0+40.0+1.0+0.0+0.0+0.5%0Aw+576+128+624+224+0%0Av+576+128+464+256+0+0+40.0+1.0+0.0+0.0+0.5%0A