EM1-6 Direct Current Circuits PDF

Electricity and Magnetism Topic 1.6: Direct-Current Circuit Resistors in Series Resistors in Parallel Kirchhoff’s Rules Based from Sears and Zemansky’s University Physics With Modern Physics (13th ed) Equivalent Resistance Resistance of a single resistor...

Electricity and Magnetism Topic 1.6: Direct-Current Circuit Resistors in Series Resistors in Parallel Kirchhoff’s Rules Based from Sears and Zemansky’s University Physics With Modern Physics (13th ed) Equivalent Resistance Resistance of a single resistor 𝑅𝑒𝑞 = 𝑉𝑎𝑏 /𝐼 We may find a single resistance that is equivalent to the resistance of strings of connected resistors. We have encountered these connections in Capacitors But in circuits, we are more interested in current than in charges. Resistors in Series and Parallel Resistors in Series Current is the same in all resistors in series. Potential difference Vab is the sum of individual potential differences. 𝑉𝑎𝑥 = 𝐼𝑅1 ; 𝑉𝑥𝑦 = 𝐼𝑅2 ; 𝑉𝑦𝑏 = 𝐼𝑅3 𝑉𝑎𝑏 = 𝑉𝑎𝑥 + 𝑉𝑥𝑦 + 𝑉𝑦𝑏 + ⋯ Equivalent Resistance for Resistors in Series 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 + ⋯ The equivalent resistance of any number of resistor in series equals the sum of their individual resistances. Req is greater than any individual resistance. Resistors in Series and Parallel Resistors in Parallel Potential difference is the same across all resistors in parallel. Total current is the sum of all the current. 𝑉𝑎𝑏 = 𝐼1 𝑅1 = 𝐼2 𝑅2 = 𝐼3 𝑅3 𝐼 = 𝐼1 + 𝐼2 + 𝐼3 + ⋯ 1 1 1 1 = + + + … Equivalent Resistance for Resistors in Parallel 𝑅𝑒𝑞 𝑅1 𝑅2 𝑅3 The reciprocal of equivalent resistance of any number of resistor in parallel equals the sum of the reciprocal of their individual resistances. Req is less than any individual resistance. Resistors in Series and Parallel Find the equivalent resistance of the network and the current along each resistor. 1 1 1 The equivalent resistance of the = + network is 6 Ω. 𝑅𝑒𝑞 𝑒𝑞 𝑅1 𝑅2 Total current is 𝑉𝑎𝑏 𝑉𝑎𝑏 18𝑉 1 1 1 0.5 𝑅𝑒𝑞 = →𝐼= = =3𝐴 𝐼 𝑅𝑒𝑞 6Ω = + = 𝑅𝑒𝑞1 = 2 Ω 𝑅𝑒𝑞1 6 3 Ω Current along 4-Ω resistor is Since 4Ω and 2Ω a𝑟𝑒 𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠: 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 also 3 A. 𝑅𝑒𝑞 = 4 Ω + 2 Ω = 6 Ω Resistors in Series and Parallel 3𝐴 = 𝐼3Ω + 𝐼6Ω 𝑉𝑐𝑏 6 𝐼3Ω = = =2𝐴 3Ω 3 𝑉𝑐𝑏 6 𝐼6Ω = = =1𝐴 6Ω 6 𝑉𝑎𝑐 = 𝐼 4Ω = 12 𝑉 If we add I 6Ω and I3Ω, it will add up to 3A. Current along 4-Ω resistor is also 3 A. Hence 𝑉𝑐𝑏 = 𝑉𝑎𝑏 − 𝑉𝑎𝑐 = 6𝑉 Resistors in Series and Parallel Kirchhoff’s Rules Below are some of the networks that cannot be reduced to simple series-parallel combinations of resistor. For these networks, Kirchhoff’s Rules are used. Kirchhoff’s Rules Junction – is a point in a circuit where three or more conductors meet Loop – is any closed conducting path Kirchhoff’s Rules are based on conservation of charges. Junction Rule, valid at any junction At any junction, incoming current is equal Σ𝐼 = 0 to outgoing current. Loop Rule, valid for any closed loop For any loop, potential differences is 0 Σ𝑉 = 0 (including EMF sources and resistive elements. Kirchhoff’s Rules Kirchhoff’s Rules The circuit shown above contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit, (b) potential difference Vab, and c) the power output of the emf of each battery. Kirchhoff’s Rules Since there are no junctions, we don’t need Junction rule. Σ𝑉 = 0 −𝐼 4Ω − 4𝑉 − 𝐼 7Ω + 12𝑉 − 𝐼 2Ω − 𝐼 3Ω = 0 8𝑉 𝐼= = 0.5 𝐴 16Ω B) To find Vab, we start at b and add V as we go toward a 𝑉𝑎𝑏 = 0.5𝐴 7Ω + 4𝑉 + 0.5𝐴 4Ω = 9.5 Point a is at high Kirchhoff’s Rules B) To find Vab, we start at b and add V as we go toward a 𝑉𝑎𝑏 = 0.5𝐴 7Ω + 4𝑉 + 0.5𝐴 4Ω = 9.5𝑉 Point a is at 9.5 higher potential at b. All term are positive because each represent an increase in potential as we go from b to a Kirchhoff’s Rules C)The power outputs of the emf of the 12-V and 4-V batteries are: 𝑃12𝑉 = 𝜉𝐼 = 12𝑉 𝑥 0.5𝐴 = 6 𝑊 𝑃4𝑉 = 𝜉𝐼 = −4𝑉 𝑥 0.5𝐴 = −2 𝑊 The negative sign in 4-V battery is due to the current running from higher to lower potential side of the battery. Negative power means we are storing energy in that battery (4-V battery). Kirchhoff’s Rules E1. In the circuit shown, find a) current in resistor R, b) the resistance R, c) the unknown emf, d) the circuit is broken at x, what is the current in resistor R? a) current in resistor R: 2.00 A b) the resistance R: 5.00 Ω c) the unknown emf: 42.0 V d) the circuit is broken at x, what is the current in resistor R? 3.50 A