Direct-current circuit/Kirchhoff's law problem

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The discussion focuses on solving a circuit problem using Kirchhoff's laws, specifically addressing the current and resistance in a given circuit. The user successfully calculated the current in resistor R as 2 A, the resistance R as 4.5 Ω, and the unknown emf ε as 42 V. However, there is confusion regarding part (d), where breaking the circuit at point x alters the current distribution, leading to an incorrect assumption that the current remains 2 A. To resolve this, it is advised to redraw the circuit without the broken branch and recalculate the current based on the new configuration. Understanding the impact of breaking the circuit is crucial for accurate calculations.
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Homework Statement


"In the circuit shown in the figure below, ε1 = 27.0 V, R1 = 6.00 Ω, and R2 = 3.00 Ω."
https://www.webassign.net/yf/27-29alt.gif

(a) Find the current in resistor R.
(b) Find the resistance R.
(c) Find the unknown emf ε.
(d) If the circuit is broken at point x, what is the current in resistor R?

I need help with understanding part d, please!

Homework Equations


I = V/R

The Attempt at a Solution


I1 + I2 - I3 = 0

ε - I2R2 - I1R2 = 0
ε1 - ε + I1R1 - I3R3 = 0

(a) Find the current in resistor R.
2 A
(b) Find the resistance R.
4.5 Ω
(c) Find the unknown emf ε.
42 V
(d) If the circuit is broken at point x, what is the current in resistor R?
2 A (wrong)

What am I doing wrong? Wouldn't the current in resistor R then be 9 V/4.5 Ω = 2 A?
 
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When you break the circuit at x it removes that branch from the circuit. It will no longer conduct current. This will change the distribution of currents and voltages in the remainder of the circuit.

Assume that component values don't change, so the value of R that you found previously remains the same. Draw the "new" circuit accordingly and determine the current.
 
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