# can u have voltage with with no current

Discussion in 'Electricians' Talk' started by brumsparky, Dec 31, 2007.

1. can u have voltage with with no current

2. Yes.

3. dormouse i new to elecrical science so forgive me ansers like this u cant find in wiki or other sites

so can u xplain how and under what circunstances pls

4. so can u xplain how and under what circunstances pls

Open circuit Voltage...

5. But!!!

(A small current still flows when you measure it..) 6. (A small current still flows when you measure it..) you should know..it's small enough

7. dormouse i new to elecrical science so forgive me
ansers like this u cant find in wiki or other sites

so can u xplain how and under what circunstances pls

Sorry.

When you buy a new battery, a voltage exists across the terminals ready to supply a current when you put it in a torch & switch on, for instance. A bit like the pressure in an aerosol paint can ready to spray when you press the button.

Wit a few exceptions, electrics are easier to understand as plumbing with electricity instesd of water.

8. That'll teach me not to use the spell check.

9. It's 5 to 4 in the morning and I've got Voltage but no current!

10. when i worked on 24V DC control systems you could have a situation were a voltmeter would measure 24v around a light circuit but the lights didnt come on.
This we called a dry joint and the only way to test it was to use a tester that loaded the circuit( test lamp).
The dry joint lets you read 24v but stops current flowing.
My electronics is a bit rusty , but thats my understanding of it.

11. Bit like an Electronic Test Meter and an Electro-mechanical one! 12. Bum Park

The simplest way to envisage this in a practical way is to consider a humble zinc-carbon AA battery. You have a potential of circa 1.5V (say Vbatt) at the terminals when not loaded.

Then apply a load (lets assume a simple resistance) then there will be a relatively small voltage drop due to the cell's internal resistance (say Vcell), a relatively small voltage drop due to the connecting circuit (say Vconnect) between the cell and the voltage across the load itself (say Vload). Vbatt=VcellVconnectVload

Regards

13. ### Mr. HandyandyScrewfix Select

Bum Park

The simplest way to envisage this in a practical way
is to consider a humble zinc-carbon AA battery. You
have a potential of circa 1.5V (say Vbatt) at the

Then apply a load (lets assume a simple resistance)
then there will be a relatively small voltage drop
due to the cell's internal resistance (say Vcell), a
relatively small voltage drop due to the connecting
circuit (say Vconnect) between the cell and the

Regards

Or, a 9.6v drill battery can read say 9.2v, but can be flat as a fart.

Mr. Handyandy - really

14. Or, a 9.6v drill battery can read say 9.2v, but can
be flat as a fart.

2v.

Mr. Handyandy - really

Whats that got todo with the OP's question ?

15. Physically, its impossible to have no current if you have voltage, as under Ohms Law, V=IR, making current 0 would make V=0 as well, and R would become infinite.

Practically, however, as I becomes very very low, V can exist at normal levels, only the power in the system would be very low too. This is how certain batteries show regular voltage, but no capacity.

You can only properly test a battery with a load test.

16. As has already been pointed out, Ohm's law of V=IR (Voltage = Current x Resistance) defines the relationship between voltage and current.

So, if you have a voltage of 100v across a 200 ohm resistance, current I = V/R = 0.5 amps flowing through that resistor.

However if the resistance is high enough (say 100 Mega ohm), then in the same scenario only 1 micro amp will flow, which is so small as to be negligible for most purposes. This might be the case where something is open circuit, but an &#147;insulator&#148; somewhere in the circuit is conducting slightly.

All voltage supplies have an inherent resistance (usually called the "internal resistance") whether from a battery or the mains. While an AA battery might read 1.5v using a multi-meter (which has a very high internal resistance when set to measure voltage), the voltage across the battery will drop when a load is applied to it.

I&#146;d like to expand on the &#147;Vbatt=VcellVconnectVload&#148; statement made previously. Assume that a 20 ohm load (a torch bulb) is connected to a &#147;flat&#148; 1.5v battery and the battery voltage drops to 1v. This must mean that a current of 1.5/20 or 75mA is flowing. The batteries internal resistance accounts for to 0.5v drop, and it must be V/R or 6.66 ohms. If the battery was fully charged its internal resistance would be much less, with very little voltage drop inside the battery.

17. Whoops, in my last post I got it slightly wrong. The battery's external voltage is only 1V. So the current flowing through the torch bulb (a 20 ohm resistor) must be 50mA. Since the battery contains a 1.5v source, there must be 0.5v lost through internal resistance as only 1V appears at the terminals/ Which means the battery's internal resistance is 0.5v/50mA=10ohms.