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- Using the rules for parallel circuits and Ohmslaw, solve for the missing values. ETE1E2E3E4ITl1I2l33.2AI4RT3.582R116R210R3R420PTP1P2P3P4Part II: The potential difference across different elements of a circuit resistance in series 1. Using the circuit shown below, set R1 and R2 as shown in the chart below. For each pair of resistances in the chart below, record the drop in voltage, or potential, across R1 (Figure a), drop in voltage, or potential, across R2 (Figure b), then record it across both resistors (Figure c). R, R, E, E, R, Figure a Figure b R, E, E, R, Figure c them Resistance Values, Voltage Drops, and Curfent Data Values R, (ohms) (ohms) R2 V, (volts) 4 1-716 V 1.47 V V2 (volts) 0.268 V V12 (volts) | (amps) 26 1.984 1.978 V 65.6 MA 22 0.531 V 65.6 MA 18 1-978 V 12 0.920 V L.182 V 0796 V 65.6 mA 1.980 V 1.981 V 14 16 1.062 V 65.6 mA 10 20 0.657V 1.325 V 24 1.985 V 65.7 mA 0- 398 V 1. 587 V 65.6 mA please help mewith Question 2 and 3 no calculation needed just a faugafh combination from the voltage and current numbers for that combination. Can this be developed 2. From the data above, explore the relationship…Please break this down for me as much as possible, im struggling to grasp kirchhoff's, thank you.
- Part A What must the emf E in the figure (Figure 1) be in order for the current through the 7.00 N resistor to be 1.65 A ? Each emf source has negligible internal resistance. Express your answer with the appropriate units. E = Value Units Submit Request Answer Figure Provide Feedback Next > 24.0 V Ţ §7.00 $2.00 3.00 NA resistance of 10 ohms and C=0.1F is connected in series with a 12V battery. Find the charge at any time and the charge current 0.4 seconds after switch is closed.Design a simple circuit using Ohm’s law. In the circuit you designed, explain how you could reduce the dissipated power by one-half.
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