Please help me with all parts of the question Solve for I (in A) and use the values given in the figure (Include the sign of the value in your answer.):E - E2(1)I =A%3D.R1+ R2because the batteries in the figure haveFinalize The negative sign for I indicates that the direction of the current is opposite the assumed direction. The emfs in the numerator ---Select---opposite polarities. The resistances in the denominator add because the two resistors are in series.EXERCISEIn the circuit in the example, suppose each battery has an internal resistance of 0.960 N.Hint(a) What will be the current (in A) in the circuit? Give only the absolute value of the current.A(b) What power (in W) is delivered to each resistor, including each battery's internal resistance?P1%3DP2%3DPintW%3D(c) What is the total power (in W) supplied to all the resistors by the two batteries?W A Single-Loop CircuitA single-loop circuit contains two resistors and two batteries as shown in the figure. (Neglect the internal resistances of the batteries.) Find the current in the circuit if R, = 6.00 N and R,= 12.0 Q.A series circuit containing two batteries andtwo resistors, where the polarities of thebatteries are in opposition.Ej = 6.0 V%3DIR2R1Ez = 12 VSOLUTIONConceptualize The figure shows the polarities of the batteries and a guess at the direction of the current. The 12 V battery is the ---Select--- 0 of the two, so the current should becounterclockwise. Therefore, we expect our guess for the direction of the current to be wrong, but we will continue and see how this incorrect guess is represented by our final answer.---Select--- O junctions in this single-loopCategorize We do not need Kirchhoff's rules to analyze this simple circuit, but let's use them anyway simply to see how they are applied. There arecircuit; therefore, the current is the same in all elements.Analyze(Use the following as necessary: I, R,, and R,. Do not substitute numerical values; use variables only.)Let's assume the current is clockwise as shown in the figure. Traversing the circuit in the clockwise direction, starting at a, we see that a → b represents a potential difference of +E,, b → crepresents a potential difference of AV, = -IR,,c→d represents a potential difference of –E2, and d → a represents a potential difference of:AV2%3DApply Kirchhoff's loop rule to the single loop in the circuit:= 0SaV = 0 → E, – IR - E2 -

Given value --- R1 = 6 ohm . R2 = 12 ohm. We have to find--- I ? P1 ? P2 ? Pint ? Ptotal ?

Hint (a) What will be the current (in A) in the circuit? Give only the absolute value of the current. A (b) What power (in W) is delivered to each resistor, including each battery's internal resistance? P1 Pint (c) What is the total power (in W) supplied to all the resistors by the two batteries?

Hint (a) What will be the current (in A) in the circuit? Give only the absolute value of the current. A (b) What power (in W) is delivered to each resistor, including each battery's internal resistance? P1 Pint (c) What is the total power (in W) supplied to all the resistors by the two batteries?

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter20: Electric Current, Resistance, And Ohm's Law

Section: Chapter Questions

Problem 7PE: (a) A defibrillator sends a 6.00-A current through the chest of a patient by applying a 10,000-V...

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