Hello people. Im not very good with electrical circuits and i was wondering if someone out there could take a few mins to help me with this problem. In the attached cicuit diagram i need to find the value of the first resistor. I hope i drew it clear enough to understand. When the switch is closed the voltmeter shows 8 volts, but when the switch is open the switch it reads 8 volts. If someone could be kind enough to explain a solution to me i would be really thankful. Thanks
Not knowing where you are in your study of electronics, I'd recommend you read DC Circuits. The circuit you've drawn is a series circuit. You haven't shared with us your thoughts on the problem.
Can you deduce the value of the battery from the information given? Once you know the battery voltage the resistor value will be clear.
The resistor must be very small compared to 10 Ohms. Can you replace the 10 Ohm resistor with a smaller resistor?
Just curious, but it looks like to me that it is normal to get the same voltage when the switch is open or closed because of where he is taking his voltage reading from. As I see it, with the switch open or closed he will be reading the voltage after it is dropped across the first resistor. This voltage reading doesn't tell us enough information since we need atleast two of three variables to solve E=IR Please tell me if I am correct in my deduction Papabravo or aac.
My reasoning is as follows. 1. With the switch open, the high impedance volt meter reads 8 volts. Only an insignificant amount of current is flowing from the battery through the resistor and the voltmeter. For practical purposes this means the open circuit voltage of the battery is 8 volts. 2. With the switch closed which means that 800 mA of current is flowing (8V/10 ohms). If the first resistor has any resistance at all then it would have a drop across it equal to 800 mA * R1. 3. We have two equations and two unknowns and the only solution that is consistent with both equations is: Battery = 8 VDC R1 = 0 ohms
The circuit as shown with the switch open draws no current, therefore you will see the voltage as 8 volts. There is no current flowing other than a minuscule amount through the voltmeter. If you close the switch and still read 8 volts then there is still no current flowing. Therefore the second resistor has infinite resistance (or be the same as open). If the second resistor were 0-ohms (a short in essence) then the meter would read 0-volts as it would be shorted. V= I*R 8v= 0 * 0 open switch. if you read 7volts then you would be able to figure out the current. you would have a 1 volt drop when the switch was closed (unless the switch were bad and wasn't closing) The second resistor is assumed to be 10 ohms then we have a voltage drop of 1 volt across it. So I= E/R 1v/10 Ohms = .1 amps in a closed circuit the current flows through the entire circuit. What you need to do is move the voltmeter across the first resistor and measure its voltage drop. when you do that then you can use ohms law for resistance. If you found out that you had a 1/2 volt drop across the first resistor then you could figure out the resistance. R = E/I so .5V/.1 amps = 5 ohms. This is not the correct answer to your problem so you need to find out both voltage drops, by using a volt meter across the battery, resistor 1 , switch, and resistor 2. I'll bet that the switch is bad and no current flows. This may be a problem in using troubleshooting methods using ohms law, so stick in there. This may be a form of mental gymnastics to find out if you know what your doing and you are paying attention to the course material. have fun.