Given:
Initially, the resistance of a cobblestone bridge equals
[tex]R_1 = 200 \ \Omega \\\\ R_2=400 \ \Omega \\\\R_3 = 500 \ \Omega \\\\ R_4 =600\ \Omega \\\\[/tex]
voltage at input
The bridge's output is:
ACB is connected to a 5V supply.
(regular voltage division)
Similarly, at point D, the voltage is
[tex]\to V_D=5 \times \frac{600}{1100}= 2.72 \ v \\\\ \to V_{out}= V_c-V_D = 3.33-2.72 = 0.6\ V = 0.61\ V\\\\[/tex]
for circuit outlined above:
[tex]\to R_1= 250\ \Omega[/tex]
then
[tex]\to V_{RI} = 5\times \frac{250}{250+400}= \frac{2.50\times 5}{250+400}= 1.923\ V\\\\[/tex]
Learn more:
brainly.com/question/19473968
