Chapter 19 DC Circuits EMF and Terminal Voltage • Devices which transfer one type of energy to electrical energy are called the sources of electromotive force (EMF). • The potential difference between terminals of such source when no current is flowing to the external circuit is called the emf of the source ( Ɛ). Remember, A battery is not a constant current source but can be assumed to be a constant voltage source. When electron is flowing within the battery, it faces some resistance due to the molecules present on it path. Hence, it is fair enough to say that every battery has its own resistance, which we call the Internal resistance (r). Hence if we use a multimeter to measure the terminal voltage, i.e., voltage across the battery terminals, we would never measure the emf of the cell. = − Unless otherwise stated, the internal resistance is considered to be negligible. Problem 1: From figure below, calculate current, terminal voltage, power dissipated in ‘R’ and in ‘r’. Resistors in Series Connection - Same current flows through each resistors. - The sum of the voltage across each of the resistor is equal to the total voltage of the battery. = 1 + 2 + 3 = 1 + 2 + 3 For the equivalent circuit, Ohm’s Law yields: = Hence, comparing the two equations, we get = + + Equivalent Circuit Resistors in Parallel Combination - Resistors are connected so that the current splits = 1 + 2 + 3 - Voltage across each resistor is the same. Hence, = 1 1 = 2 2 = 3 3 = 1 + 2 + 3 = 1 1 + 1 2 + Also, in the equivalent circuit, I = Thus, 1 1 1 1 = + + 1 2 3 1 3 …. (1) … (2) Problem 2: Determine (a) equivalent resistance and (b) the voltage across each resistor, and (c) current through each resistor. Solution: Get calculator ready to help me !! a) 357 + 990 Ohm b) 8.8, 3.2 V c) 8.91, 4.7,4.2 mA Remember: Series - Current same, voltage add up Parallel - Voltage same, current add up Problem 3: Find net resistance of the circuit. Find current through each resistor if R= 3.25k- Ohm and V=12V. Also what is the potential difference between point A and B. Solution: a) 5/8R + R b) I1=I2= 0.284mA I3= 0.568 mA I4= 0.852mA I5= 1.42mA I6=2.27mA c) 1.85V Kirchhoff’s Law: - Used to solve circuits with multiple loops and voltage source. Law of current: At any junction, the sum of all currents entering must be equal to sum of all currents leaving. For the circuit above: 1 + 2 = 3 + 4 + 5 Law of Voltage: The sum of the changes in potential around any closed loop of a circuit must be zero. Problem 4: Calculate the magnitude and direction of the currents in each resistor. Approach: 1) Choose current direction with some thoughts 2) Write current equation at the junctions 3) Write voltage equation for the loop Answer: 120 Ω: 0.25 A 56 Ω: 0.16 A 110 Ω: 0.090 A 74 Ω: 0.25 A 25 Ω: 0.090 A Capacitors in Series and Parallel = Parallel: • Total charge gets split so that = 1 + 2 + 3 = 1 + 2 + 3 So, = + + Series: • Same charge flows • Voltage splits = 1 + 2 + 3 = + + 1 2 3 So, = + + Problem 5: Calculate (a) Voltage across each capacitor (b) charge on each capacitor. Approach: (a) Q = CV - Voltage across C3 = 21 V - Voltage across C12 (C1 series to C2)= 21 V. Since Q is the same in series: C1V1=C2V2 with V= V1 + V2 So, 3V1 = 4(21-V1) V1 = 12 V , and V2 = 9V. (b) Charge across C3, Q3= C3V=2x21 Charge across C1, Q1 = C1V1 = 3x12 Charge across C2, Q2 = C2V2 = 4X9 RC Circuits: - Car windshield, Pacemaker, camera flashes Charging: As switch is turned on, electron moves from -ve terminal of the battery to one of the capacitor plate, which incudes +ve charge on the other plate. A –ve free charge on the second plate moves to the +ve terminal of the battery. The charge continues to deposit on the capacitor until Ɛ = Vc. = 1 − − Since Q= CV, above equation can be written as: − = 0 1 − = (Time Constant) Time constant is the measure of how quickly capacitor gets charged. - e is exponential function, it’s value is 2.71. For t = RC, we can see Vc = 0.63. Also, Ɛ = Vc + VR . So, VR = − . Above figure shows this variation. RC Circuit Discharging: Read Yourself !!! Problem 6: Total Resistance is 15 K Ω, emf 24 V. If the time constant is measured to be 18 μs, calculate (a) total capacitance and (b) time it takes for voltage across the resistor to reach 16V after the switch is closed. Solution: = . 18 x 10-6 = 15 x 1000 x C So, C = 1.2 x 10-9 F. --------------------------------------------- − During Charging: VR = . − 18×10−6 Hence, 16 = 24 Divide both sides by 24, and take log: ln(0.66) = -t/18x10-6 -0.415 x 18x 10-6 = -t Hence, t = 7.48 x 10-6 s.
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