When two 100 Ohm resistors are connected in parallel, we first need to find the equivalent resistance of the first two resistors. The formula for the equivalent resistance (\(R_{eq}\)) of resistors in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
For the first two 100 Ohm resistors:
\[ \frac{1}{R_{eq}} = \frac{1}{100} + \frac{1}{100} = \frac{2}{100} = \frac{1}{50} \]
Thus:
\[ R_{eq} = 50 , \text{Ohm} \]
When these two resistors are connected to a 100 V power supply, the voltage across each resistor remains the same as the power supply voltage (100 V) because they are in parallel. Both resistors will have 100 V across them.
Now, when you add two more 100 Ohm resistors in parallel with these, the total equivalent resistance will be recalculated. The new total resistance is determined as follows:
The equivalent resistance of the second pair of 100 Ohm resistors is also:
\[ \frac{1}{R_{eq2}} = \frac{1}{100} + \frac{1}{100} = \frac{2}{100} = \frac{1}{50} \]
So, \(R_{eq2} = 50 , \text{Ohm}\) for the second pair as well.
Now, we can find the new total equivalent resistance of the four resistors:
\[ \frac{1}{R_{total}} = \frac{1}{R_{eq1}} + \frac{1}{R_{eq2}} = \frac{1}{50} + \frac{1}{50} = \frac{2}{50} = \frac{1}{25} \]
Thus,
\[ R_{total} = 25 , \text{Ohm} \]
Since the power supply is still 100 V, the voltage across each of the original and the new sets of resistors remains at 100 V due to them all being in parallel. Therefore, the voltage across the new resistors, like the first two, is:
\[ \text{Voltage across the new resistors} = 100 , V \]
Hence, the answer is 100 V.