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Current conservation, power and fundamental circuits Review: I = σA V/d = V / R where A is a cross section area of a wire, d is length. V is the voltage across the wire. We call the quantity characterizing the material conductivity σ. Inverse of σ is called resistivity ρ, ρ = 1/σ R= ρ d/A

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Resistance Valid for “ohmic” devices mainly metallic conductors at constant temperature.

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Conservation of current

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Q3. What is the magnitude of the current flowing through the conductor at the bottom?

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Circuit elements Elements with negligible resistance---Wires. We assume that there is no potential difference across a wire.

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Electrical Hazards Feel: 1mA pain: few mA deadly: over 70mA

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Electrical Power P = VQ/t = VI Unit: Watt (W) = VA kW, MW, GW

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Electrical power on the ohmic device P = VI V = RI (Ohm’s Law) P = V 2 /R = I 2 R

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Q2. The 60 W bulb is working at 120V. What is the current flowing through the bulb?

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Q3. The 60 W bulb is working at 120V. What is the resistance of the bulb (in ohms)?

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Fuses and switch

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Q4. The voltage across the nail is 2V at 25A. What is the resistance of the nail (in Ohms)?

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Q5. The voltage across the nail is 2V at 25A. What is the power dissipated in the nail?

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Kirchhoff's rules At any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction. ( Conservation of charge) The Sum of the changes in potential around any closed path of the circuit must be zero

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Combinations of resistors Resistors in series: V = V 1 +V 2 +V 3 = IR 1 + IR 2 + IR 3 R eq = V/I = R 1 + R 2 + R 3 V R1R1 R2R2 R3R3

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Combinations of resistors Resistors in parallel I = I 1 +I 2 +I 3 = V/ R 1 + V/ R 2 + V/R 3 I = V/R eq 1/R eq = 1/ R 1 + 1/ R 2 + 1/R 3 R3R3 V R2R2 R1R1

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Combinations of resistors Resistors in parallel 1/R eq = 1/ R 1 + 1/ R 2 + 1/R 3 R eq = R 1 ∙R 2 ∙R 3 / (R 1 ∙R 2 + R 2 ∙R 3 + R 1 ∙R 3 ) Notice R eq is smaller than all of them! R3R3 V R2R2 R1R1

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