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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 RA + RB 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? Known Unknown R1 = 10 Ω R = ?? R2 = 15 Ω I = ??? R3 = 5 Ω

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**A) What is the equivalent resistance in the circuit?**

R = RA + RB + RC 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 RA + RB + RC 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 / RA +RB VB = IRB VB = IRB = V x RB RA +RB VB = VRB

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**A) What is the current in the circuit? **

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? Known Unknown V source = 45.0 V I = ?? RA = 47 Ω VA = ?? RB = 82 Ω VB = ??

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**A) What is the current in the circuit?**

Find equivalent resistance R = RA + RB I = V source / R = V source / RA +RB I = V 47 Ω +82 Ω I = A

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**Find the voltage drop of each resistor**

VA = IRA = (0.349)(47 Ω) = 16 V VB = IRB = (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 / RA +RB I = V 39 Ω +82 Ω = 0.372 VB = IRB = (0.372)(82Ω) = 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 = IA + IB + IC …**

So then…. V = V + V + V … RA RB RC 1/R = (1/RA) + (1/RB) + (1/RC) …

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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/RA) + (1/RB) + (1/RC)**

= 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 IA = V / RA IB = V / RB **

= 90.0 V / 60.0 Ω = 1.50 A IB = V / RB = 90.0 V / 30.0 Ω = 3.00 A IC = V / RC = 90.0 V / 20.0 Ω = 4.50 A

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Concept Review The _______________is the same everywhere in a simple series circuit The equivalent ______________of a series circuit is the ___________of.

Concept Review The _______________is the same everywhere in a simple series circuit The equivalent ______________of a series circuit is the ___________of.

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