a 12.0 olm resistor, a 6.0 olm resistor, and a 8.0 olm resistor are connected in serieds to a 20 volt battery?
363388
The total resistance is 12 + 8 + 6 = 26 ohms. In series the total resistance is the sum of individual resistors. Apply ohm's law V=IR, I = 20/26 Amperes. The voltage drop across each resistor is found by multiplying the current with the resistance of each resistor.
225854
12.5 is fine. the battery is 12 volts.
313707
Assuming you are using US household electricity (120VAC)\nP=V^2/R   R=V^2/P = (120)^2/1875= 7.68 ohms which is really small.
644810
Your circuit has 6 resitance R1, R2, R3, R4, R5, R6 and correspond currents are: I1, I2, I3, I4, I5, I6 and voltage are: U1, U2, U3, U4, U5, U6.\nYou can make many kind of circuit(direct, parallel or both) depend on total voltage as well as resistance.
964185
I sense that you have no clue what you're doing. Tell me I'm wrong.\n\nFor DC circuits, Ohm's law says: V = I * R, voltage equals current times resistance.\n\nYour circuit will consist of a battery (voltage source) and a load (resistance) plus the wires to connect them and the meter(s).\n\nIdeally you will test one battery at a time. That way the load will be identical (including the small resistance of the wires).\n\nBatteries typically have two ratings: voltage and capacity.\nVoltage is measured in volts and this decreases as the battery discharges. Capacity is measured in amp hours (current multiplied by time) which specifies how long the battery can deliver a current.\n\nAssuming the circuit resistance is constant (reasonable assumption for your purposes), when you connect the battery to the load, the voltage (independent variable) will cause a current (dependent variable) to flow. As the battery discharges, the voltage will decrease, and the current will correspondingly decrease. You will need a watch or timer as well. Time zero will be when you first connect the battery. Measure the battery voltage at that instant and at intervals. Depending on the load and how fast the battery discharges should determine how frequently you take measurements.\n\nmsg me when you have the data if you need help analyzing it.
572940
A voltmeter consists of a galvanometer(to detect current) and a high resistance(usually thousands) connected in series. It measures the voltage across two points. Theorectically a voltmeter should have infinite resistance.
784102
16. C 150 Ohms not Watts\n\n17. A\n\n18. D Current Divider Rule in a parallel Circuit\n\n19. C\n\n20. A  6V time 0.04A = 0.24 Watts
418101
You need the internal resistance. Then based on the amperage due to load you can figure the I^2R loss in watts for the battery.\n\nThe internal resistance can be determined by adding a resistive load to the battery and monitoring the voltage. The point at which the voltage starts to drop equals the internal resistance of the battery.
645621
http://www.swa.org/site/hhw/batteries2/hhw_batteries2.htm
372456
To measure ohms, DCV, ACV, Amps or what????
1078584
There are some solution available at http://www.batteryspace.com/index.asp?PageAction=VIEWCATS&Category=885
969011
You might care to explain what some of the letters are, not all of them have much to do with electronics.\n\nR I guess here is resistance, but what of l m x and s?  Is l for inductance (I think not here), what of s? seconds?\n\nSome of these (all ?) are simpley equations (eg: c=q/E solve for E) would apply whatever C, ! and E stood for.\n\nIn this case if you gave C an arbitar value or 2, Q an arbitrary value of 6 ad E an arbitary value of 3, you would see that C (2) = q (6) / E (2).   So E would be Q/C\n\nThis is usually the I=v/r fourmal where I=current, v=voltage and r=resistance. R=v/I is also the same but the other way around. V=IR is valid also.\n\ny=mx+b?  what electrical formla is that? Again it is easy to solve but I can see nothing to with electronics here. Again try assigning sinple numbers to the letters so that the equation works - then you can re-arrange the sum so it still works:\n\nGive m, x and b a different value: 3, 4+ 5\n\nNow 3*4 +5  = 17\nso Y=17.\n\nNow you can play about with re-arranging your sum:   \n\n5 = 17   -    3 * 4 \n\n5 represents b, 17 represents Y, 3 represents m and 4 represents x\n\nSo swapping numbers for letters:    b=y-mx
254527
1.85 ohms. The first answer is correct
797986
This is a very simple circuit.\nhttp://www.rason.org/Projects/fmkeyer/fmkeyer.htm
616731
The equivalent resistance of the parallel resistances R3, R4, R5 is given by\n\n1/Re = 1/5 + 1/10 + 1/ 15. =  55/150\nRe     = 2.73 ohm.\nThe equivalent resistance of the series resistances =R1+R2+Re =32.73 ohm\nTotal resistance =32.73 ohm.\n\nCurrent total = voltage /total resistance = 10 /32.73 ohm = 0.305 A.\n\nVolts of R1 = current x resistance = 0.305 x 10 = 3.05 V.\nVolts of R2 = current x resistance = 0.305 x 20 = 6.11 V.\nVolts of Re   = current x resistance = 0.305 x 2.73 = 0.83265 V.\n\nVolts of R3= Volts of R4= Volts of R5= Volts of Re =0.83265 V.\n\nCurrent through R1 = Current through R2 = Current through Re =  0.305 A\n\nCurrent through R3 = volts of R3/ R3 = 0.83265/5 =0.16653A\nCurrent through R4 = volts of R4/ R4 = 0.83265/10 =0.083265A\nCurrent through R5  = volts of R5/ R5 = 0.83265/15 =0.05551A
821830
You have to know with what kind of battery your starting.
225576
TTL: 1 - 10k Ohms works, CMOS: 10 - 20 KOhms