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REFERENCE: https://drive.google.com/file/d/1T4NFEWABIUicQ1HJ2-2SfyPylUF6M0wt/view?usp=sharing please answer thank you so much. i will give a high rate. please write legibly so that i can read it. follow also

REFERENCE: https://drive.google.com/file/d/1T4NFEWABIUicQ1HJ2-2SfyPylUF6M0wt/view?usp=sharing

please answer thank you so much. i will give a high rate. please write legibly so that i can read it. follow also the formula given in the module (which is in the reference) give the given and the formula used in every items.

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What's More Resistors and Kirchhoff's Rule Problems Instructions. Solve the given problem. Identify the given and the unknown and then show your complete solution. Write your answer in your answer sheet. Bases for rating: 5 points - correct answer with comprehensive solutions 3 points - answer was provided without a comprehensive solution 1 point - tries to answer 1. A 12.0 V storage battery is connected to three resistors, 6.7 Q, 15.3 Q and 21.6 Q, respectively. The resistors are joined in series. a. Calculate the equivalent resistance. b. What is the current in the circuit? 2. A 4.0 2 resistor, an 8.0 2 resistor, and a 12.0 Q resistor are connected in series with a 24.0 V battery. a. Calculate the equivalent resistance. b. Calculate the current in the circuit. c. What is the current in each resistor? 3. Find the current in and potential difference across each of the resistors in the following circuits: a. A 150 Q and a 180 Q resistor wired in series with a 12 V source. b. A 150 Q and a 180 Q resistor wired in parallel with a 12 V source. 4. A string of 35 miniature lights decorative lights is wired in series. If it draws 0.020 A of current when it is connected to a 120.0 emf source, what is the resistance of each miniature bulb? 5.a. Find the currents in the circuit shown: Two loops, two batteries: Find the currents. See illustration below. A 12 9V 1000 / Loop 1 ER R Loop 2 13 15V 1000 1000 I1 B 1. Apply the junction rule to point A: I1 - 12- 13 = 0 2. Apply the loop rule to loop 1 (let R = 100 0): 15V - IR - I,R = 03. Apply the loop rule to loop 2 (let R = 100 0): -9V - L2R + 13R = 0 4. Solve for I1 = ?, 12 = ?, 13 = ? Note that /2 is negative. This means that its direction is opposite to that shown in the circuit diagram. 5. Suppose the polarity of the 9V battery is reversed. What are the currents in this case? I, = ?, Iz = ?, Is=

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