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23-1 Simple Circuits

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Series Circuits Brightness of lamps is depended upon the ______________ in a circuit When all current travels in all of the devices, said to be in Series B/c current travels through all, current is same through out I = V source R A + R B Equivalent Resistance Sum of all individual resistance

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Problems A 10 ohm resistor, 15 ohm resistor, and a 5 ohm resistor are connected in series across a 90 V battery. A) What is the equivalent resistance in the circuit? B) What is the current in the circuit? KnownUnknown R 1 = 10 Ω R = ?? R 2 = 15 Ω I = ??? R 3 = 5 Ω

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A) What is the equivalent resistance in the circuit? R = R A + R B + R C R = 10 Ω + 15 Ω + 5 Ω R = 30 Ω

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B) What is the current in the circuit? I = V source R A + R B + R C I = 90 V 10 Ω + 15 Ω + 5 Ω I = 3 A

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Voltage Drop Voltage Divider A Series circuit used to produce a voltage source of desired magnitude from a higher voltage battery Solve by rearranging the R = V/I V = IR R is the equivalent resistance and V is the potential drop

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Suppose you have a 9-V battery but need a 5-V potential source. I = V/R = V / R A +R B V B = IR B V B = IR B = V x R B R A +R B V B = VR B R A +R B

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Two Resistors, 47 Ω and 82 Ω are connected in series across a 45.0 V Battery A) What is the current in the circuit? B) What is the voltage drop in each resistor? C) The 47 Ω resistor is replaced by a 39 Ω resistor. Will the current increase, decrease, or stay the same? D) What will happen to the voltage drop across the 82 Ω resistor? KnownUnknown V source = 45.0 VI = ?? R A = 47 Ω V A = ?? R B = 82 Ω V B = ??

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A) What is the current in the circuit? Find equivalent resistance R = R A + R B I = V source / R = V source / R A +R B I = 45.0 V 47 Ω +82 Ω I = A

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Find the voltage drop of each resistor V A = IR A = (0.349)(47 Ω ) = 16 V V B = IR B = (0.349)(82 Ω ) = 29 V

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Calculate Current again using RA and determine new voltage drop I = V source / R = V source / R A +R B I = 45.0 V 39 Ω +82 Ω = V B = IR B = (0.372)(82 Ω ) = 31 V

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A 9.0 V battery and two resistors, 400 ohms and 500 ohms, are connected as a voltage divider. What is the voltage across the 500 ohm resistor?

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Parallel Circuits A circuit in which there are several current paths Total current is the sum of the currents through each path and the potential difference across each path is the same Current through path depends on each resistor

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Parallel Circuits When resistance is removed, current through other resistors does not change, only the total current through the circuit

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Parallel Circuits Finding Equivalent resistance I = I A + I B + I C … So then…. V = V + V + V … R A R B R C 1/R = (1/R A ) + (1/R B ) + (1/R C ) …

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Three resistors, 60.0 Ω, 30.0 Ω, and 20.0 Ω, are connected in parallel across a 90.0 V battery A) Find the current through each branch of the circuit B) Find the equivalent resistance of the circuit C) Find the current through the battery

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2) Find the equivalent resistance 1/R = (1/R A ) + (1/R B ) + (1/R C ) 1/R = (1/60 Ω ) + (1/30 Ω ) + (1/20 Ω ) = Ω -1 R = 10.0 Ω 3) Find the Total Current I = V/R = 90.0 V / 10.0 Ω = 9.00 A So we want the total current equal to the individual currents added together

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Find the current through each branch I A = V / R A = 90.0 V / 60.0 Ω = 1.50 A I B = V / R B = 90.0 V / 30.0 Ω = 3.00 A I C = V / R C = 90.0 V / 20.0 Ω = 4.50 A

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