# Solving Electrical Resistances & Calculating Work & Mass: Help Needed!

• ritwik06
In summary, the conversation covers topics such as resistors in parallel and series, work done on springs and potential energy, and finding relevant material for further learning. The specific question being addressed is finding a missing value in a series of resistors, the amount of work done on a spring when compressed, and the size of a wooden cube that just dips in water. The conversation ends with the person thanking the expert for providing helpful information.
ritwik06
1. Three resistances of value 1,2 and 5 have to be connected in such a way that all are used once. Which value cannot be obtained?
1) 17/3 2) 17/4 3) 17/6 4)17/7

I know formulas for parallel and interconnected resistance. My teacher tells me that by looking at the options I can solve this. Pleasse help. I am not able to make out.

2. In stretching a spring through 4 cm, 2 J work is done. If it compressed by 6 cm then amount of work will be?

According to me its 3j but my book gives it 4.5 J. How come?

3. A wooden cube (densit 0.60 g/cc) just dips inside the water if pressed with a force of 10 N. What must be the size of cube? (g= 10m/s^2)

I have no idea.

You should show some efforts in order to get help.

Hint: regarding 2), what is the relation between work and the change of potential energy in a spring?

You should show some efforts in order to get help.

Hint: regarding 2), what is the relation between work and the change of potential energy in a spring?

I am sorry sir. I have problems. I am a student of class 9th. I take part in all the olympiads to learn new things. I have nothing to read except my own class textbooks. I just use the internet. Just imagine as to how can I think about something about which I don't have the slightest idea. I need help from you all (if its not a burden). Please try solving my questions and tell me about the topics and books or sites in which I could find relevant material.

regards

First of all, you don't have to call me 'sir', not that I don't appreciate your politeness though.

Second, for number 2), you should find this link useful: http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html" .

Last edited by a moderator:
First of all, you don't have to call me 'sir', not that I don't appreciate your politeness though.

Second, for number 2), you should find this link useful: http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html" .

Thank you very very much for your help. You gave me exactly what I needed. Thanks a million times!

Last edited by a moderator:

## 1. How do I calculate electrical resistance?

Electrical resistance can be calculated using Ohm's Law, which states that resistance (R) is equal to the voltage (V) divided by the current (I): R = V/I. You can also use the formula R = ρl/A, where ρ is the resistivity of the material, l is the length of the material, and A is the cross-sectional area of the material.

## 2. What is the relationship between electrical resistance and work?

Electrical resistance is directly proportional to the amount of work done in an electrical circuit. The higher the resistance, the more work is required to push the electrical current through the circuit. This is because resistance causes a voltage drop, which is the energy that is converted into work.

## 3. How do I calculate work in an electrical circuit?

Work in an electrical circuit can be calculated using the formula W = V x Q, where W is the work done, V is the voltage, and Q is the charge. You can also use the formula W = I² x R x t, where I is the current, R is the resistance, and t is the time.

## 4. How do I calculate mass in an electrical circuit?

Mass in an electrical circuit can be calculated using the formula m = QV²/2E, where m is the mass, Q is the charge, V is the voltage, and E is the energy. This formula is based on the principle of energy conservation, which states that energy cannot be created or destroyed, only transformed.

## 5. Can you provide an example of solving electrical resistances?

Sure! Let's say you have a circuit with a voltage of 12 volts and a current of 3 amps. You also know that the length of the material is 5 meters and the cross-sectional area is 2 square meters. Using the formula R = ρl/A, we can calculate the resistance: R = (12 V)/(3 A) = 4 Ω. This means that for every 1 volt of electrical potential, 4 amps of current will flow through the circuit.

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