Electrical Engineering - Series-Parallel Circuits - Discussion

It is given that in which of the following options voltmeter presents a minimum load means here the voltmeter is behaving as a load. For more and more voltages the coil in the voltmeter has low value.

For example take a voltage divider. If the voltage setting is less than full voltage appears across the load.

In the same way when coil resistance is low full voltage will be appeared across load that means the voltmeter is not much acting as a load. So 1000v is correct. If in options any value which is given more than that then it will be correct.

For voltmeter, the maximum load current and maximum load Resistor is fixed.

i.e. the maximum voltage is fixed. Read or measured voltage in excess of this are dropped across an added series resistor.

So, Vtest = Vmeter + Vseries.

= Vmeter + Imeter * Rseries.

Since Vmeter and Imeter are fixed.

=> Vtest proportional to Rseries.

So, increasing Vtest will reqire increasing Rseries.

Also Rtotal = Rmeter + Rseries.

The higher the rating of V meter, the higher its internal resistance. Since V meter is connected in parallel in circuit, to have minimum loading effect the V meter should have high internal resistance so that minimum amount of circuit current can pass through it. Thus, among the four options, the highest rating V meter is 1000V and thus it will result in minimum withdrawal of circuit current.

By ohms law we know that current is directly proportional to voltage and inversely proportional to resistance. Adding load means adding resistance we all know that when resistance is high, current is low when resistance is low current is high.

So here resistance is low means current is high which is directly proportional to voltage so answer is max in given options 1000 is max.

@All.

A voltmeter's setting is made in terms of Ohm/V.

So, for sensitive circuits, the voltmeter will draw minimum current when its resistance is maximum which will be for V=1000V available in options.

For example, if Voltmeter sensitivity is 1000 Ohm/V then for 1000 Volt, its resistance will be 1000 * 1000 = 1000000 Ohm and it will draw negligible current.
That's it.

@Yamini.

No its not possible to measure the load with voltage which is present by a voltmeter. But its true.

Voltage is inversely proportional to load on the system i.e. as load increases voltage decreases. But we can't measure individual voltmeter we need flowing current or opposing resistance.

Voltage is inversely proportional to load on the system-i.e as load increases voltage decreases due to v-drops hence they are inversely related. So minimum load pertains to maximum voltage among the given options.

Here is the concept of relationship between load res & Voltage.

Since load is lesser as per question so the voltage value will be the higher one (having less voltage drop on account of less load in the ckt).

@Yamini.

Voltmeter present means voltmeter shown minimum load on the circuit its means current flow in circuit is low then voltmeter shown volt in maximum value. So we say voltage range 1000v.

@Puja.

The voltage is inversely proportional to load. So the minimum load has maximum voltage so the answer is max voltage that is 1000V.