Calculating Resistance and Power in a Simple Circuit

[warnexus]

Homework Statement

Homework Equations

current = (emf)/(resistance)

power = (Capacitance)(Voltage)/ (2)

The Attempt at a Solution

i might be misunderstanding the question but when in part A when it says 5V across it, does it mean if the emf was 5 V instead of the given 12 V?

current = (5V)/(270 ohm)
current = .04 Amperes

current is same in a capacitors

.04 = 5V(Resistance_2)
Resistance = 120 ohms

 

[SammyS]

Homework Statement

Homework Equations

current = (emf)/(resistance)

power = (Capacitance)(Voltage)/ (2)

The Attempt at a Solution

i might be misunderstanding the question but when in part A when it says 5V across it, does it mean if the emf was 5 V instead of the given 12 V?

current = (5V)/(270 ohm)
current = .04 Amperes

current is same in a capacitors

.04 = 5V(Resistance_2)
Resistance = 120 ohms

No. It does not mean that you replace the 12V with 5V.

Do you know Kirchhoff's Circuit Laws?

In this series circuit, the sum of the voltage drops across the resistors is equal to the emf of the battery.

According to Ohm's Law, the voltage drop across a resistor of resistance, R, which has current, I, flowing through it is I∙R .

 

[warnexus]

charges gain energy at emf and lose energy in resistors. what you stated previously would mean the voltage drop is 12 V. i thought you would only use kirchhoff's law if there is more than one emf or if the circuit is connected in a complex way

let see at emf there is a gain of 12 V. part a says 5 V across so that's a 5 V voltage drop in R_2. that would also mean a 7 V drop in R_1. since a gain of 12 V added with (-12 V drop) must sum to zero based on the kirchhoff's law

 

[SammyS]

charges gain energy at emf and lose energy in resistors. what you stated previously would mean the voltage drop is 12 V

A voltage drop of 5V across R₂ means there is a voltage drop of 7V across R₁ because the source emf is 12V .

 

[warnexus]

A voltage drop of 5V across R₂ means there is a voltage drop of 7V across R₁ because the source emf is 12V .

yes that is right. apparently I need to draw a CW(Clock wise loop to represent the state of emf) in this region of the circuit

i noticed I also need to find power dissipation which are in joules. which would mean I cannot used the stored energy equation. how would I go about finding power dissipation?

 

[SammyS]

yes that is right. apparently I need to draw a CW(Clock wise loop to represent the state of emf) in this region of the circuit

i noticed I also need to find power dissipation which are in joules. which would mean I cannot used the stored energy equation. how would I go about finding power dissipation?

Do you have a textbook or notes? This is very basic stuff.